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Digitized  by  the  Internet  Archive 

in  2008  with  funding  from 

Microsoft  Corporation 


http://www.archive.org/details/elementarytextboOOsnydrich 


ELEMENTARY   TEXTBOOK 


ON    THE 


CALCULUS 


BY 
VIRGIL  SNYDER,   Ph.D. 

AND 

JOHN    IRWIN   HUTCHINSON,   Ph.D 
Of  Cornell  University 


NEW  YORK  •:•  CINCINNATI  •:•  CHICAGO 

AMERICAN    BOOK    COMPANY 


THE    MODERN    MATHEMATICAL    SERIES. 
Lucien  Augustus  Wait, 

{Senior  Professor  of  Mathematics  in  Cornell  University ,) 

General  Editor. 


This  series  includes  the  following  works  : 
BRIEF  ANALYTIC  GEOMETRY.    By  J.  H.  Tanner  and  Joseph  Allen. 
ELEMENTARY  ANALYTIC  GEOMETRY.    By  J.  H.  Tanner  and  Joseph 

Allen. 
DIFFERENTIAL  CALCULUS.    By  James  McMahon  and  Virgil  Snyder. 
INTEGRAL  CALCULUS.     By  D.  A.  Murray. 
DIFFERENTIAL  AND  INTEGRAL  CALCULUS.    By  Virgil  Snyder  and 

J.  I.  Hutchinson. 
ELEMENTARY    TEXTBOOK   ON    THE    CALCULUS.      By  Virgil  Snyder 

and  J.  I.  Hutchinson. 


HIGH  SCHOOL   ALGEBRA.     By  J.  H.  Tanner. 
ELEMENTARY  ALGEBRA.    By  J.  H.  Tanner. 
ELEMENTARY  GEOMETRY.    By  James  McMahon. 


COPYRIGttl',    1912,    BY 

AMERICAN  BOOK  COMPANY 


EL.    CALCULUS. 
W.    P.    1 


Q  fV3  03 


53/ 


PREFACE 


The  present  volume  is  the  outgrowth  of  the  requirements 
for  students  in  engineering  and  science  in  Cornell  University, 
for  whom  a  somewhat  brief  but  adequate  introduction  to  the 
Calculus  is  prescribed. 

The  guiding  principle  in  the  selection  and  presentation  of 
the  topics  in  the  following  pages  has  been  the  ever  increasing 
pressure  on  the  present-day  curriculum,  especially  in  applied 
science,  to  limit  the  study  of  mathematics  to  a  minimum  of 
time  and  to  the  topics  that  are  deemed  of  most  immediate  use 
to  the  professional  course  for  which  it  is  preparatory. 

To  what  extent  it  is  wise  and  justifiable  to  yield  to  this 
pressure  it  is  not  our  purpose  to  discuss.  But  the  constantly 
accumulating  details  in  every  pure  and  applied  science  makes 
this  attitude  a  very  natural  one  towards  mathematics,  as  well 
as  towards  several  other  subjects  which  are  subsidiary  to  the 
main  object  of  the  given  course. 

This  desire  to  curtail  mathematical  training  is  strikingly 
evidenced  by  the  numerous  recent  books  treating  of  Calculus 
for  engineers,  for  chemists,  or  for  various  other  professional 
students.  Such  books  have  no  doubt  served  a  useful  purpose 
in  various  ways.  But  we  are  of  the  opinion  that,  in  spite  of 
the  unquestioned  advantages  of  learning  a  new  method  by 
means  of  its  application  to  a  specific  field,  a  student  would 
ordinarily  acquire  too  vague  and  inaccurate  a  command  of  the 
fundamental  ideas  of  the  Calculus  by  this  one-sided  presenta- 
tion.    While  a  suitable  illustration  may  clear  up  the  difficulties 

3 

262792 


4/(  •  ;/f  ;'l  :*•/;•  \    *  .,  'breface 

of  an  abstract  theory,  too  constant  a  dwelling  among  applica- 
tions alone,  especially  from  one  point  of  view,  is  quite  as  likely 
to  prevent  the  learner  from  grasping  the  real  significance  of 
a  vital  principle. 

In  recognition  of  the  demand  just  referred  to,  we  have  made 
special  effort  to  present  the  Calculus  in  as  simple  and  direct 
a  form  as  possible  consistent  with  accuracy  and  thoroughness. 
Among  the  different  features  of  our  treatment,  we  may  single 
out  the  following  for  notice. 

The  derivative  is  presented  rigorously  as  a  limit.  This  does 
not  seem  to  be  a  difficult  idea  for  the  student  to  grasp,  espe- 
cially when  introduced  by  its  geometrical  interpretation  as 
the  slope  of  the  line  tangent  to  the  graph  of  the  given  func- 
tion. For  the  student  has  already  become  familiar  with  this 
notion  in  Analytic  Geometry,  and  will  easily  see  that  the 
analytic  method  is  virtually  equivalent  to  a  particular  case  of 
the  process  of  differentiation  employed  in  the  Calculus. 

In  order  to  stimulate  the  student's  interest,  easy  applications 
of  the  Differential  Calculus  to  maxima  and  minima,  tangents 
and  normals,  inflexions,  asymptotes,  and  curve  tracing  have 
been  introduced  as  soon  as  the  formal  processes  of  differentia- 
tion have  been  developed.  These  are  followed  by  a  discussion 
of  functions  of  two  or  more  independent  variables,  before  the 
more  difficult  subject  of  infinite  series  is  introduced. 

In  the  chapter  on  expansion,  no  previous  knowledge  of  series 
is  assumed,  but  conditions  for  convergence  are  discussed,  and 
the  criteria  for  determining  the  interval  of  convergence  of  those 
series  that  are  usually  met  with  in  practice  are  derived. 

A  chapter  on  the  evaluation  of  indeterminate  forms  and 
three  chapters  on  geometric  applications  furnish  ample  illus- 


PREFACE  5 

tration  of  the  uses  of  infinite  series  in  a  wide  range  of 
problems. 

By  reason  of  its  significance  in  applications,  it  does  not  seem 
advisable  to  omit  the  important  principle  of  rates.  Arising 
out  of  the  familiar  notion  of  velocity,  it  affords  an  early  glimpse 
into  applications  of  the  Calculus  to  Mechanics  and  Physics. 
We  do  not  propose  to  make  the  Calculus  a  treatise  on  Mechanics, 
as  seems  to  be  the  tendency  with  some  writers;  but  a  final 
chapter  on  applications  to  such  topics  of  Mechanics  as  are  easy 
to  comprehend  at  this  stage  is  thought  advisable  and  sufficient. 
Especially  in  treating  of  center  of  gravity,  the  formulas  have 
been  derived  in  detail,  first  for  n  particles,  and  then,  by  a  limit- 
ing process,  for  a  continuous  mass.  This  was  considered  the 
more  desirable,  as  textbooks  in  applied  mathematics  frequently 
lack  in  rigor  in  discussing  the  transition  from  discrete  particles 
to  a  continuous  mass.  Besides,  the  derivation  of  these  formu- 
las affords  a  very  good  application  of  the  idea  of  the  definite 
integral  as  the  limit  of  a  sum.  This  idea  has  been  freely  and 
consistently  used  in  the  derivation  of  all  applied  formulas  in 
the  Integral  Calculus.  However,  as  the  formula  for  the  length 
of  arc  in  polar  coordinates  is  especially  difficult  of  derivation 
by  this  method,  we  have  deduced  it  from  the  corresponding 
formula  for  rectangular  coordinates  by  a  transformation  of  the 
variable  of  integration. 

In-order  to  make  the  number  of  new  ideas  as  few  as  possible, 
the  notions  of  infinitesimals  and  orders  of  infinitesimals  have 
been  postponed  to  the  last  article  on  Duhamel's  principle.  This 
principle  seems  to  flow  naturally  and  easily  from  the  need  of 
completing  the  proof  of  the  formulas  for  center  of  gravity. 
The  teacher  may  omit  this  article,  but  its  presence  should  at 


6  PREFACE 

least  serve  the  important  end  of  calling  the  attention  of  the 
student  to  the  fact  that  there  is  something  yet  to  be  done  in 
order  to  make  the  derivations  complete. 

Some  teachers  will  undoubtedly  prefer  to  do  a  minimum 
amount  of  work  in  formal  integration  and  use  integral  tables  in 
the  chapters  on  the  applications.  For  such  the  first  chapter  of 
the  Integral  Calculus  might  suffice  for  drill  in  pure  integration. 
The  problems  in  this  chapter  are  numerous,  and,  for  the  most 
part,  quite  easy,  and  should  furnish  the  student  a  ready  insight 
into  the  essential  principles  of  integration. 

The  characteristic  features  of  the  books  on  the  Calculus 
previously  published  in  this  series  have  been  retained.  The 
extensive  use  of  these  books  by  others,  and  a  searching  yearly 
test  in  our  own  classroom  experience  convince  us  that  any  far- 
reaching  change  could  not  be  undertaken  without  endangering 
the  merits  of  the  book.  The  changes  that  have  been  made  are 
either  in  the  nature  of  a  slight  rearrangement,  or  of  the  addi- 
tion of  new  illustrative  material,  particularly  in  the  applications. 

We  wish  to  acknowledge  our  indebtedness  to  our  colleagues, 
who  have  added  many  helpful  suggestions ;  to  Professor  I.  P. 
Church,  of  the  College  of  Civil  Engineering,  for  a  number  of 
very  useful  problems  in  applications  of  integration  (See  Exs. 
14-18,  pp.  318-320,  and  Exs.  6-7,  pp.  323-324),  and  particu- 
larly to  Professor  James  McMahon,  who  has  carefully  read  all 
the  manuscript,  assisted  throughout  in  the  proof  reading,  and 
made  many  improvements  in  the  text. 


CONTENTS 

DIFFERENTIAL   CALCULUS 
CHAPTER  I 

Fundamental  Principles 

ARTICLE  PAGE 

1.  Elementary  definitions 15 

2.  Illustration  :  Slope  of  a  tangent  to  a  curve  .         .         .         .16 

3.  Fundamental  theorems  concerning  limits 17 

4.  Continuity  of  functions 19 

5.  Comparison  of  simultaneous  increments  of  two  related  variables  20 

6.  Definition  of  a  derivative 21 

7.  Process  of  differentiation 22 

8.  Differentiation  of  a  function  of  a  function          ....  23 

CHAPTER  II 
Differentiation  of  the  Elementary  Forms 


9. 

Differentiation  of  the  product  of  a  constant  and  a  variable 

25 

10. 

Differentiation  of  a  sum      . 

26 

11. 

Differentiation  of  a  product 

27 

12. 

Differentiation  of  a  quotient 

28 

13. 

Differentiation  of  a  commensurable  power  of  a  function    . 

29 

14. 

Differentiation  of  implicit  functions 

33 

15. 

Elementary  transcendental  functions          .... 

34 

16. 

Differentiation  of  loga  x  and  loga  u 

34 

17. 

Differentiation  of  the  simple  exponential  function 

36 

18. 

Differentiation  of  an  incommensurable  power     . 

37 

19. 

Limit  of  ^—  as  6  approaches  0 

38 

20. 

Differentiation  of  sin  u 

39 

21. 

Differentiation  of  cos  u 

40 

22. 

Differentiation  of  tan  u 

40 

23. 

Differentiation  of  sin-1  u 

42 

24. 

Table  of  fundamental  forms 

44 

CONTENTS 


CHAPTER  III 

Successive  Differentiation 

ARTICLE  PAGE 

25.  Definition  of  the  nth  derivative 47 

26.  Expression  for  the  nth  derivative  in  certain  cases      ...      49 

CHAPTER  IV 
Maxima  and  Minima 

27.  Increasing  and  decreasing  functions 51 

28.  Test  for  determining  intervals  of  increasing  and  decreasing       .  51 

29.  Turning  values  of  a  function 53 

30.  Critical  values  of  the  variable 55 

31.  Method  of  determining  whether  <f>'(x)  changes  its  sign  in  pass- 

ing through  zero  or  infinity      .......       55 

32.  Second  method  of  determining  whether  <p'(x)  changes  its  sign 

in  passing  through  zero  ...  .....       57 

33.  The  maxima  and  minima  of  any  continuous  function   occur 

alternately .59 

34.  Simplifications  that  do  not  alter  critical  values   ....       59 

35.  Geometric  problems  in  maxima  and  minima      ....      60 

CHAPTER  V 
Rates  and  Differentials 

36.  Rates.     Time  as  independent  variable 68 

37.  Abbreviated  notation  for  rates 72 

38.  Differentials  often  substituted  for  rates 74 

39.  Theorem  of  mean  value 74 

CHAPTER  VI 

Differential  of  an  Area,  Arc,  Volume,  and 
Surface  of  Revolution 

40.  Differential  of  an  area 78 

41.  Differential  of  an  arc 79 

42.  Trigonometric  meaning  of  — ,  — .80 

dx    dy 

43.  Differential  of  the  volume  of  a  surface  of  revolution  ...      81 


CONTENTS 


ARTICLE 

44.  Differential  of  a  surface  of  revolution 

45.  Differential  of  arc  in  polar  coordinates 

46.  Differential  of  area  in  polar  coordinates 


81 
82 
83 


CHAPTER  VII 
Applications  to  Curve  Tracing 

47.  Equation  of  tangent  and  normal  .... 

48.  Length  of  tangent,  normal,  subtangent,  and  subnormal 

49.  Concavity  upward  and  downward 
60.  Algebraic  test  for  positive  and  negative  bending 

51.  Concavity  and  convexity  toward  the  axis    . 

52.  Hyperbolic  and  parabolic  branches     . 

53.  Definition  of  a  rectilinear  asymptote  . 


85 
85 
89 
90 
94 
95 
96 


Determination  of  Asymptotes 

54.  Method  of  limiting  intercepts 96 

55.  Method  of  inspection.     Infinite  ordinates,  asymptotes  parallel 

to  axes 97 

56.  Method  of  substitution.     Oblique  asymptotes     ....       99 

57.  Number  of  asymptotes 102 

Polar  Coordinates 

58.  Meaning  of  p  — 104 

dp 

59.  Relation  between  ^  and  p— 105 

dx  dp 

60.  Length  of  tangent,  normal,  polar  subtangent,  and  polar  sub- 

normal      105 


CHAPTER  VIII 


Differentiation  of  Function 

61.  Definition  of  continuity 

62.  Partial  differentiation 

63.  Total  differential 

64.  Total  derivative  .  • 

65.  Differentiation  of  implicit  functions 

66.  Geometric  interpretation     . 

67.  Successive  partial  differentiation 

68.  Order  of  differentiation  indifferent 


s  of  Two  Variables 


109 
110 
112 
115 
116 
118 
122 
122 


10  CONTENTS 

CHAPTER  IX 

Change  of  Variable 

article  page 

69.  Interchange  of  dependent  and  independent  variables          .         .  124 

70.  Change  of  the  dependent  variable 125 

71.  Change  of  the  independent  variable 126 

72.  Simultaneous  changes  of  dependent  and  of  independent  variables  126 

CHAPTER  X 
Expansion  of  Functions 

73.  Convergence  and  divergence  of  series 132 

74.  General  test  for  convergence ,  133 

75.  Interval  of  convergence 138 

70.    Remainder  after  n  terms ,     .  140 

77.  Maclaurin's  expansion  of  a  function  in  a  power  series        .        .141 

78.  Taylor's  series 148 

79.  Rolle's  theorem 150 

80.  Form  of  remainder  in  Maclaurin's  series 150 

81.  Another  expression  for  the  remainder 153 

CHAPTER   XI 

t 

Indeterminate  Forms 

82.  Definition  of  an  indeterminate  form »  157 

83.  Indeterminate  forms  may  have  determinate  values     .         .         .  158 

84.  Evaluation  by  development 160 

85.  Evaluation  by  differentiation       .......  161 

86.  Evaluation  of  the  indeterminate  form  g- 165 

CHAPTER  XII 
Contact  and  Curvature 

87.  Order  of  contact 167 

88.  Number  of  conditions  implied  by  contact 168 

89.  Contact  of  odd  and  of  even  order 169 

90.  Circle  of  curvature 172 

91 .  Length  of  radius  of  curvature  ;  coordinates  of  center  of  curvature  1 72 

92.  Limiting  intersection  of  normals 174 

93.  Direction  of  radius  of  curvature          .         .         .  '               •         •  175 


CONTENTS 


11 


ARTICLE  PAGE 

94.  Total  curvature  of  a  given  arc  ;  average  curvature    .         .         .  176 

95.  Measure  of  curvature  at  a  given  point 177 

.  96.   Curvature  of  an  arc  of  a  circle 178 

97.  Curvature  of  osculating  circle 178 

98.  Direct  derivation  of  the  expressions  for  k  and  B  in  polar  co- 

ordinates   180 

EvOLUTES   AND    INVOLUTES 

99.  Definition  of  an  evolute 182 

100.   Properties  of  the  evolute 184 

CHAPTER   XIII 
Singular  Points 


101.  Definition  of  a  singular  point 

102.  Determination  of  singular  points  of  algebraic  curves 

103.  Multiple  points 

104.  Cusps         ......... 

105.  Conjugate  points 


191 
191 
193 
194 
197 


CHAPTER  XIV 
Envelopes 


106.  Family  of  curves 

107.  Envelope  of  a  family  of  curves 

108.  The  envelope  touches  every  curve  of  the  family 

109.  Envelope  of  normals  of  a  given  curve 

110.  Two  parameters,  one  equation  of  condition 


200 
201 
202 
203 
204 


12  CONTENTS 

INTEGRAL   CALCULUS 
CHAPTER  I 

General  Principled  oe  Integration 

ARTICLE  PAGE 

111.  The  fundamental  problem 209 

112.  Integration  by  inspection 210 

113.  The  fundamental  formulas  of  integration  ....  211 

114.  Certain  general  principles 212 

115.  Integration  by  parts 216 

116.  Integration  by  substitution 219 

117.  Additional  standard  forms 222 

118.  Integrals  of  the  forms  C(^x  +  B)dx  and  C  (Ax  +  B)dx       #     »M 

J  ax2  +  bx  +  c         J  ^/ax-2  +  bx  +  c 

119.  Integrals  of  the  forms  f — and 

J  (Ax  +  B)  Vax2  +  bx  +  c 

dx 


5 


(Ax  +  B)2  Vax2  +  bx  +  c 


121.  Decomposition  of  rational  fractions 

122.  Case  I.     Factors  of  the  first  degree,  none  repeated 

123.  Case  II.     Factors  of  the  first  degree,  some  repeated 

124.  Case  II J.     Occurrence  of  quadratic  factors,  none  repeated 

125.  Case  IV.     Occurrence  of  quadratic  factors,  some  repeated 

126.  General  theorem 


226 


CHAPTER  II 
120.   Reduction  Formulas 229 

CHAPTER  III 

Integration  of  Rational  Fractions 


238 
240 
241 
243 
245 
247 


CHAPTER   IV 

Integration  by  Rationalization 

127.    Integration  of  functions  containing  the  irrationality  y/ax  +  b      248 


128.  Integration  of  expressions  containing  Vax2  +  bx  +  c        .         .     249 

129.  The  substitution  V±  t1  ±  k2  -z 253 


CONTENTS 


13 


CHAPTER  V 

Integration  of  Trigonometric  Functions 


ARTICLE 

130.   Integration  by  substitution 


181.  Integration  of  (  sec2na;  dx,    \  csc2»x  dx 

132.  Integration  of  I  secTOx  tan'2n+1x  dx,   j  cscmx  coV^+h;  dx 

133.  Integration  of  i  tan"x  dx,   \  cotnx  dx 

134.  Integration  of  (  sin'"x  cosnx  dx 

135.  Integration  of  f — ,   f- 

J  a  +  b  cos  nx   J  a 

k 


dx 
b  sin  nx 


dx 


+  b  sin  nx  -f  c  cos  nx 
136.   Integration  of  I  eax  sin  wxcJx,   je^cos  nxdx 


PAGE 

255 
255 

257 

258 

260 

264 
266 


CHAPTER  VI 

Integration  as  a  Summation.     Areas 


137.  Areas 268 

138.  Expression  of  area  as  a  definite  integral 270 

139.  Generalization  of  the  area  formula 273 

140.  Certain  properties  of  definite  integrals 274 

141.  Maclaurin's  formula 276 

142.  Remarks  on  the  area  formula -  277 

143.  Precautions  to  be  observed  in  evaluating  definite  integrals       .  283 

144.  Calculation  of  area  when  x  and  y  are  expressible  in  terms  of  a 

third  variable 289 

145.  Areas  in  polar  coordinates 291 

146.  Approximate  integration.     The  trapezoidal  rule       .         .         .  292 

147.  Simpson's  rule 294 

148.  The  limit  of  error  in  approximate  integration  ....  295 


14 


CONTENTS 


CHAPTER   VII 

Geometrical  Applications 

ARTICLE  PAGE 

149.  Volumes  by  single  integration 298 

150.  Volume  of  solid  of  revolution 302 

151.  Lengths  of  curves.     Rectangular  coordinates    ....  306 

152.  Lengths  of  curves.     Polar  coordinates 309 

153.  Measurement  of  arcs  by  the  aid  of  parametric  representation  .  310 

154.  Area  of  surface  of  revolution 312 

155.  Various  geometrical  problems  leading  to  integration                 .  315 


CHAPTER  VIII 
Successive  Integration 


156.  Functions  of  a  single  variable    . 

157.  Integration  of  functions  of  several  variables 

158.  Integration  of  a  total  differential 

159.  Multiple  integrals       .         .         . 

160.  Definite  multiple  integrals 

161.  Plane  areas  by  double  integration     . 

162.  Volumes 


321 
324 
326 
328 
329 
330 
334 


CHAPTER  IX 

Some  Applications  of  Integral  Calculus  to 
Problems  of  Mechanics 


163. 

Liquid  pressure  on  a  plane  vertical  wall   . 

.      .         .     338 

164. 

Center  of  gravity 

.     340 

165. 

Moment  of  inertia     . 

.     346 

166. 

Duhamel's  theorem  . 

.     348 

Trigonometric  Formulas 

.     352 

Logarithmic  Formulas 

.     353 

DIFFERENTIAL   CALCULUS 


^XKc 


CHAPTER   I 

FUNDAMENTAL  PRINCIPLES 

1.  Elementary  definitions.  A  constant  number  is  one  that 
retains  the  same  value  throughout  an  investigation  in  which  it 
occurs.  A  variable  number  is  one  that  changes  from  one  value  to 
another  during  an  investigation.  If  the  variation  of  a  number 
can  be  assigned  at  will,  the  variable  is  called  independent;  if 
the  value  of  one  number  is  determined  when  that  of  another  is 
known,  the  former  is  called  a  dependent  variable.  The  depend- 
ent variable  is  called  also  a  function  of  the  independent  variable. 


E.g.,  3  x2,  4vx  —  1,  cos  x,  are  all  functions  of  x. 

Functions  of  one  variable  x  will  be  denoted  by  the  symbols 
/(#),  <f>(x),  •  ••,  which  are  read  as  "/of  x"  " <f>  of  x"  etc. ;  simi- 
larly, functions  of  two  variables,  x,  y,  will  be  denoted  by  such 
expressions  as 

f(?,y),F(x,y),  •••• 

When  a  variable  approaches  a  constant  in  such  a  way  that 
the  difference  between  the  variable  and  the  constant  may  be- 
come and  remain  smaller  than  any  fixed  number,  previously 
assigned,  the  constant  is  called  the  limit  of  the  variable. 

15 


16 


DIFFERENTIAL   CALCULUS 


2.  Illustration :  Slope  of  a  tangent  to  a  curve.  To  obtain  the 
slope  of  the  tangent  to  a  curve  at  a  point  P  upon  it,  first  take 
the  slope  of  the  line  joining  P  =  (xly  yx)  to  another  point  (x2,  y2) 
upon  the  curve,  then  determine  the  limiting  value  of  the  slope 


m 


as  the  second  point  approaches  to  coincidence  with  the  first, 
always  remaining  on  the  curve. 

Ex.  1.     Determine  the  slope  of  the  tangent  to  the  curve 


2-FM  +  fc 


at  the  point  (2,  4)  upon  it. 

Here,  x\  =  2,  y\  =  4.  Let  x2  =  2  +  h, 
yi  =  4  +  k,  where  h,  k  are  so  related  that  the 
point  (x2,  y*)  lies  on  the  curve. 

Thus  4  +  k  =  (2  +  h)\ 

or  h  =  4  A  +  A2-  (1) 


The 

slope 

m  =  y*  - 

x2  - 

-Xi 

becomes 

4+  Tc 
2  +  h 

-4 

_  2 

k 

Fig.  1 


which  from  (1)  may  be  written  in  the  form 

k 


=  4  +  h. 


(2) 


The  ratio  k :  h  measures  the  slope  of  the  line  joining  (xh  yx)  to 
(ar2,  ys) •  When  the  second  point  approaches  the  first  as  a  limiting 
position,  the  first  member  of  equation  (2)  assumes  the  indeterminate 
form  -,  but  the  second  member  approaches  the  definite  limit  4.    When 

the  two  points  approach  coincidence,  a  definite  slope  4  is  obtained, 
which  is  that  of  the  tangent  to  the  curve  y  =  x2  at  the  point  (2,  4). 

It  may  happen  that  h,  k  appear  in  both  members  of  the  equation 
which  defines  the  slope,  as  in  the  next  example. 


FUNDAMENTAL   PRINCIPLES 


17 


Fig.  2 


Ex.  2.     If  x2  +  y1  —  «2?  find  the  slope  of  the  tangent  at  the  point 
Oi>  yd- 
Since 

Xl*  +  i/!2  =  a2,  (asi  +  ny+  (t/1  +  ky  =  a\ 

hence     2  hx1  +  A2  +  2  /fr/i  +  fc2  =  0, 

from  which  -  =  —  - — — — 
h  2  ?/i  +  k 

k 
To  obtain  the  limit  of  -,   put  h,  k 
h 

each  equal  to  zevo  in  tlie  second  member. 

lim  *  =  _*!. 

h±o  h  ?/i 

This  step  is  more  fully  justified  in  the  next  article. 
This  result  agrees  with  that  obtained  by  elementary  geome- 
try.    The  slope  of  the  radius  to  the  circle  a2  +  y2  =  a2  through 

the  point  (xlf  yx)  is  — ,  and  the  slope  of  the  tangent  is  the  nega- 

tive  reciprocal  of  that  of  the  radius  to  the  point  of  tangency, 
since  the  two  lines  are  perpendicular. 

3.    Fundamental  theorems  concerning  limits.      The  following 
theorems  are  useful  in  the  processes  of  the  Calculus. 

Theorem  1.     If  a  variable  a  approaches  zero  as  a  limit,  then 
lea  will  also  approach  zero,  k  being  any  finite  constant. 

That  is,  if  a  =  0, 

then  Jca  =  0. 

For,  let  c  be  any  assigned  number.     By  hypothesis,  a  can  be- 
come less  than  -,  hence  ka  can  become  less  than  c,  the  arbi- 

k        ■  ' 

*  For  convenience,  the  symbol  =  will  be  used  to  indicate  that  a  variable 
approaches  a  constant  as  a  limit;  thus  the  symbolic  form  x  =  a  is  to  be  read 
"  the  variable  x  approaches  the  constant  a  as  a  limit." 
el.  calc  —  2 


18  DIFFERENTIAL   CALCULUS 

trary,  assigned  number,  hence  ka  approaches  zero  as  a  limit. 
(Definition  of  a  limit.) 

Theorem  2.  Given  any  finite  number  n  of  variables 
a,  (3,  y,  •••,  each  of  which  approaches  zero  as  a  limit,  then  their 
sum  will  approach  zero  as  a  limit.  For  the  sum  of  the  n 
variables  does  not  at  any  stage  numerically  exceed  n  times  the 
largest  of  them,  which  by  Theorem  1  approaches  zero. 

Theorem  3.  If  each  of  two  variables  approaches  zero  as  a 
limit,  their  product  will  approach  zero  as  a  limit.  More  gen- 
erally, if  one  variable  approaches  zero  as  a  limit,  then  its 
product  with  any  other  variable  having  a  finite  limit  will  have 
the  limit  zero,  by  Theorem  1. 

Theorem  4.  If  the  sum  of  a  finite  number  of  variables  is 
variable,  then  the  limit  of  their  sum  is  equal  to  the  sum  of 
their  limits ;  i.e., 

lim  (x  +  y  +  •  •  •)  =  lim  x  +  lim  y  + 
For,  if  x  =  a,         y  =  b,  •  •  •, 

then  x  =  a  +  a,        y  =  b  -\-  (3,  •-•, 

wherein  a  =  0,         fi  =  0,  ••  • ;    (Def .  of  limit) 

hence  x  +  y+  •••  =  (o+  &+  •••)  +  («  +  fi+  •••)> 

but  a  +  p+->-  =0,  (Th.  2) 

hence,  from  the  definition  of  a  limit, 

lim  (x  +  y  +  •••)  =  a-\-b-\-  •••  =  lim  x  -f-  lim  y  +  •••. 

Theorem  5.  If  the  product  of  a  finite  number  of  variables 
is  .variable,  then  the  limit  of  their  product  is  equal  to  the 
product  of  their  limits. 

For,  let  x  =  a  +  a,         y  =  b+($, 

wherein  a  =  0,        (3  =  0, 

so  that  lim  x  =  a,        lim  y  =  b. 


FUNDAMENTAL  PRINCIPLES  19 

Form  the  product 

xy  =  (a  +  a)(b  +  fi)  =  ab  +  «6  -f  /3a  +  «0. 
Then  lim  sc?/  =  lim  (ab  +  ab  +  (5a  +  a(5) 

=  ab  +  lim  ab  +  lim  0a  +  lim  a/3  (Th.  2) 

=  ab.  (Th.  1) 

Hence      lim  xy  =  lim  a;  •  lim  y. 

In  the  case  of  a  product  of  three  variables  x,  y,  z,  we  have 
lim  xyz  =  lim  xy  •  lim  z  (Th.  5) 

=  lim  x  lim  y  lim  3, 
and  so  on,  for  any  finite  number  of  variables. 

Theorem  6.  If  the  quotient  of  two  variables  as,  y  is  vari- 
able, then  the  limit  of  their  quotient  is  equal  to  the  quotient 
of  their  limits,  provided  these  limits  are  not  both  infinite  or 
not  both  zero. 


(Th.  5) 


y     lim  y 

4.  Continuity  of  functions.  When  an  independent  variable  x, 
in  passing  from  a  to  b,  passes  through  every  intermediate 
value,  it  is  called  a  continuous  variable.  A  function  f(x)  of  an 
independent  variable  x  is  said  to  be  continuous  at  any  value  xl} 
or  in  the  vicinity  of  xXi  when  f(x^)  is  real,  finite,  and  determi- 
nate, and  such  that  in  whatever  way  x  approaches  a^, 

From  the  definition  of  a  limit  it  follows  that  corresponding 
to  a  small  increment  of   the  variable,  the  increment  of   the 


For,  since 

X 

x  =  -y, 

y 

lim  x  =  lim  -  lim  yy 

y 

and  hence 

,.     x      lim  x 
lim  -  = 

20 


DIFFERENTIAL   CALCULUS 


function  is  also  small,  and  that  corresponding  to  any  number 
c,  previously  assigned,  another  number  8  can  be  determined, 
such  that  when  h  remains  numerically  less  than  8,  the  differ- 

is  numerically  less  than  c. 


2/,+e 


a>.+5 


Fig.  3 


Thus,  the  function  of  Fig.  3  is  continuous  between  the  values 
xx  and  xY  -f-  8,  while  the  functions  of  Fig.  4  and  Fig.  5  are  dis- 
continuous. In  the  former  of  these  two  the  function  becomes 
infinite  at  x  =  c,  while  in  the  latter  the  difference  between  the 
value  of  the  function  at  c  +  h  and  c  —  h  does  not  approach 
zero  with  h,  but  approaches  the  finite  value  AB  as  h  ap- 
proaches zero. 

When  a  function  is  continuous  for  every  value  of  x  between 
a  and  b,  it  is  said  to  be  continuous  within  the  interval  from  a 
to  b. 

5.  Comparison  of  simultaneous  increments  of  two  related  vari- 
ables. The  illustrations  of  Art.  2  suggest  the  following  general 
procedure  for  comparing  the  changes  of  two  related  variables. 
Starting  from  any  fixed  pair  of  values  x1}  y^  represented  graph- 
ically by  the  abscissa  and  ordinate  of  a  chosen  point  P  on  a 
given  curve  whose  equation  is  given,  we  change  the  values  of 


FUNDAMENTAL  PRINCIPLES 


21 


x  and  y  by  the  addition  of  small  amounts  h  and  k  respectively, 
so  chosen  that  the  new  values  xL  +  h  and  yl  +  k  shall  be  the 
coordinates  of  a  point  P2  on  the 
curve.  The  amount  h  added  to  xx 
is  called  the  increment  of  x  and  is 
entirely  arbitrary.  Likewise,  k  is 
called  the  increment  of  y ;  it  is  not 
arbitrary  but  depends  upon  the 
value  of  h ;  its  value  can  be  calcu- 
lated when  the  equation  of  the  curve 
is  given,  as  is  shown  by  equation  (1).  These  increments  are 
not  necessarily  positive.  In  the  case  of  continuous  functions,  h 
may  always  be  taken  positive.  The  sign  of  k  will  then  depend 
upon  the  function  under  consideration.     The  slope  of  the  line 

PjP2  is  then  -  and  the  slope  of  the  tangent  line  at  Pj  is  the 

limit  of  -  as  h  and  consequently  k  approach  zero. 

The  determination  of  the  limit  of  the  ratio  of  k  to  h  as  h  and 
k  approach  zero  is  the  fundamental  problem  of  the  Differential 
Calculus.  The  process  is  systematized  in  the  following  ar- 
ticles. While  the  related  variables  are  here  represented  by 
ordinate  and  abscissa  of  a  curve,  they  may  be  any  two  related 
magnitudes,  such  as  space  and  time,  or  volume  and  pressure  of 
a  gas,  etc. 

6.  Definition  of  a  derivative.  If  to  a  variable  a  small  incre- 
ment  is  given,  and  if  the  corresponding  increment  of  a  contin- 
uous function  of  the  variable  is  determined,  then  the  limit  of 
the  ratio  of  the  increment  of  the  function  to  the  increment  of 
the  variable,  when  the  latter  increment  approaches  the  limit 
zero,  is  called  the  derivative  of  the  function  as  to  the  variable. 


22  DIFFERENTIAL   CALCULUS 

k 
That  is,  the  derivative  is  the  limit  of  -  as  h  approaches  zero, 


or 


liin   (k 


For  the  purpose  of  obtaining  a  derivative  in  a  given  case  it  is 
convenient  to  express  the  process  in  terms  of  the  following  steps: 

1.  Give  a  small  increment  to  the  variable. 

2.  Compute  the  resulting  increment  of  the  function. 

3.  Divide  the  increment  of  the  function  by  the  increment  of 
the  variable. 

4.  Obtain  the  limit  of  this  quotient  as  the  increment  of  the 
variable  approaches  zero. 

7.  Process  of  differentiation.  In  the  preceding  illustrations, 
the  fixed  values  of  x  and  of  y  have  been  written  with  sub- 
scripts to  show  that  only  the  increments  h,  k  vary  during  the 
algebraic  process  of  finding  the  derivative,  also  to  emphasize 
the  fact  that  the  limit  of  the  ratio  of  the  simultaneous  incre- 
ments h,  k  depends  upon  the  particular  values  which  the 
variables  x,  y  have,  when  they  are  supposed  to  take  these  in- 
crements. With  this  understanding  the  subscripts  will  hence- 
forth be  omitted.  Moreover,  the  increments  h,  k  will,  for 
greater  distinctness,  be  denoted  by  the  symbols  Ax,  Ay,  read 
"  increment  of  x,"  "  increment  of  y." 

If  the   four  steps  of  Art.  6   are   applied   to  the  function 
y  =  <£  (x),  the  results  become 
y  +  £fy=<f>(x  +  \x), 

Ay  =  <j>(x  +  Ax)  —  <f>(x)  =  A<f>  (x), 

Ay  _<f>(x  +  Ax)  —  <j>  (x)  _  A<£  (x) 

Ax  Ax  Ax 

.       lira  -AJ  =  Km  {+(*  +  **)-  »(*)  |  =Um  Aj>^ 

A#  Ax  Ax 


'       FUNDAMENTAL   PRINCIPLES  23 

The  operation  here  indicated  is  for  brevity  denoted  by  the 

symbol  — ,  and  the  resulting  derivative  function  by  <f>'(x);  thus 
dx 


dy  _d<f>(x)  _    lim     f  <f>(x  +  Ax)-<j>(x)' 


dx         dx         Aa;  =  0 


Ax 


=  +-(*). 


The  new  symbol  -^  is  not  (at  the  present  stage  at  least)  to 
ax 

be  looked  upon  as  a  quotient  of  two  numbers  dy,  dx,  but  rather 
as  a  single  symbol  used  for  the  sake  of  brevity  in  place  of 
the  expression  "  derivative  of  y  with  regard  to  x." 

The  process  of  performing  this  indicated  operation  is  called 
the  differentiation  of  <f>  (x)  with  regard  to  x. 

EXERCISES 
Find  the  derivatives  of  the  following  functions  with  regard  to  x. 

5.  I. 

X3 

6.  xn,  n  being  a  positive  integer. 

7-2 

7. 


1. 

x2-  2x-,  2x\  3;  x. 

2. 

3x*-4:x  +  3. 

3. 

1 
4*' 

4. 

**-2  +  i. 

X2 

9.    y  =  Vx. 

10.   y  =  x~$. 

8. 


ar+1 
x 


f  1 

[Put  #2  =  x,  and  apply  the  rules.] 


8.  Differentiation  of  a  function  of  a  function.  Suppose  that  y, 
instead  of  being  given  directly  as  a  function  of  x,  is  expressed 
as  a  function  of  another  variable  u,  which  is  itself  expressed 
as  a  function  of  x.  Let  it  be  required  to  find  the  derivative 
of  y  with  regard  to  the  independent  variable  x. 

Let  y  =f(u),  in  which  u  is  a  function  of  x.  When  x  changes 
to  the  value  au  +  Aaj,  let  u  and  y,  under  the  given  relations, 


24  DIFFERENTIAL   CALCULUS 

change  to  the  values  u  +  Aw,  y  +  A?/.     Then 

A?/  _  Aw      Aw , 

Aa; —  Aw      Ax 

hence,  equating  limits  (Th.  5,  Art.  3), 

dy  _dy      da  _  df(u)      du 
dx  ~~  du     dx~    du        dx 

This  result  may  be  stated  as  follows : 

The  derivative  of  a  function  of  u  with  regard  to  x  is  equal  to 
the  product  of  the  derivative  of  the  function  with  regard  to  w,  and 
the  derivative  of  u  with  regard  to  x. 

EXERCISES 

1.  Given  v  =  3u2-l,  M  =  3x2  +  4;  find  ^- 

dy  du 

du  dx 

dx      du      dx  K  ' 

2.  Given  ?/ =3m2-4u+ 5,«  =  2x3-5;  find  ^ • 

3.  Given  y  =  -,w  =  5a;2-2x  +  4;  find  ^  • 

1  -r3      3      _    ,  rfv 

3  m2  3       a:3  aar 


CHAPTER   II 

DIFFERENTIATION  OF  THE   ELEMENTARY  FORMS 

dv 
In  recent  articles,  the  meaning  of  the  symbol  -f  was  ex- 

ctx 

plained  and  illustrated ;  and  a  method  of  expressing  its  value, 
as  a  function  of  x,  was  exemplified,  in  cases  in  which  y  was  a 
simple  algebraic  function  of  x,  by  direct  use  of  the  definition. 
This  method  is  not  always  the  most  convenient  one  in  the  dif- 
ferentiation of  more  complicated  functions. 

The  present  chapter  will  be  devoted  to  the  establishment  of 
some  general  rules  of  differentiation  which  will,  in  many  cases, 
save  the  trouble  of  going  back  to  the  definition. 

The  next  five  articles  treat  of  the  differentiation  of  algebraic 
functions  and  of  algebraic  combinations  of  other  differentiable 
functions. 

9.   Differentiation  of  the  product  of  a  constant  and  a  variable. 


Let 

y  =  cu, 

Then 

y  +  Ay  =  c(u  +  Au), 

A?/  =  c(m  +  Au)  —  cu  =  cAu, 

Ay        Au 

Ax        Ax' 

therefore 

dy        du 
dx~~    dx 

Thus 

d(cu)  _    du 
dx          dx 

25 

(1^ 


26  DIFFERENTIAL   CALCULUS 

The  derivative  of  the  product  of  a  coyistant  and  a  variable  is 
equal  to  the  constant  multiplied  by  the  derivative  of  the  variable. 

10.   Differentiation  of  a  sum. 

Let  2/==M_[_<y_W;_j_   ... 

in  which  u.  v,  w,  ••>  are  functions  of  x. 

Then  y  +  Ay  =  u  +  Au  +  v  +  Av  —  w  —  Aw  +  •  •  •, 

Ay  =  Au  +  A?;  —  Aiv  +  •  •  •, 

Ay  _  Au  ,Av_  Aw 

Ax     Ax      Ax      Ax  ' 

dy  _du     dv     dw 
dx     dx     dx      dx 

Hence        -f(u  +  v-  w+  .  ..)=f^+^-^+    ..  (2) 

doc  doc     doc     doc 

The  derivative  of  the  sum  of  a  finite  number  of  fractions  is 
equal  to  the  sum  of  their  derivatives. 

Cor.     If  y  =  u  +  c,  c  being  a  constant,  then 
y  +  Ay  =  u  +  Au  +  c, 
hence  Ay  =  Au, 

and  dy  =  du 

dx     dx 

This  last  equation  asserts  that  all  functions  which  differ 
from  each  other  only  by  an  additive  constant  have  the  same 
derivative. 

Geometrically,  the  addition  of  a  constant  has  the  effect  of 
moving  the  curve  y  =  u(x)  parallel  to  the  y-axis ;  this  opera- 
tion will  obviously  not  change  the  slope  at  points  that  have 

the  same  x. 

-c         /rtN  dy     du  ,  dc 

From  (2),  -f-  =  —  +  — ; 

dx     dx     dx 


DIFFERENTIATION  OF  THE   ELEMENTARY  FORMS      27 

but  from  the  fourth  equation  above, 

dy  _du% 
dx     dx' 

dc 
hence,  it  follows  that       —  =  0. 
dx 

The  derivative  of  a  constant  is  zero. 

If  the  number  of  functions  is  infinite,  Theorem  4  of  Art.  3  may  not 
apply;  that  is,  the  limit  of  the  sum  may  not  be  equal  to  the  sum  of 
the  limits,  and  hence  the  derivative  of  the  sum  may  not  be  equal  to 
the  sum  of  the  derivatives.  Thus  the  derivative  of  an  infinite  series 
cannot  always  be  found  by  differentiating  it  term  by  term. 

11.   Differentiation  of  a  product. 

Let  y  =  uv,  wherein  u,  v  are  both  functions  of  x. 

Then      ^=(U  +  *UW  +  *V)-UV  =  u^  +  v^  +  ^  .  to,. 
Ax  Ax  Ax         Ax      Ax 

Now  let  Aa;  approach  zero,  using  Art.  3,  Theorems  4,  5,  and 

noting  that  if  —  has  a  finite  limit,  then  the  limit  of  Avf—) 

Ax  \AxJ 

is  zero.  ' 

The  result  may  be  written  in  the  form 

d(uv)  =  udv  +  vdu  (3) 

doc  dx        doc 

The  derivative  of  the  product  of  two  functions  is  equal  to  the 
sum  of  the  products  of  the  first  factor  by  the  derivative  of  the  sec- 
ond, and  the  second  factor  by  the  derivative  of  the  first. 

This  rule  for  differentiating  a  product  of  two  functions  may 
be  stated  thus :  Differentiate  the  product,  treating  the  first 
factor  as  constant,  then  treating  the  second  factor  as  constant, 
and  add  the  two  results. 


28  DIFFERENTIAL   CALCULUS 

Cor.     To  find  the  derivative  of  the  product  of  three  functions 
uvw. 

Let  y  =  uvw. 

By  (3),  *y  =  w±(uv)+uv^ 

dx         dx  dx 


=  w(u 


dv        du\  dw 

dx        dx )  dx 


The  result  may  be  written  in  the  form 

d(uvw)=uvdw  +  vwdu  +  wudv  (4 

doc  dx  dm  dx 

By  induction  the  following  rule  is  at  once  derived : 

The  derivative  of  the  product  of  any  finite  number  of  factors  is 
equal  to  the  sum  of  the  products  obtained  by  midtiplying  the  de- 
rivative of  each  factor  by  cdl  the  other  factors. 

12.   Differentiation  of  a  quotient. 

Let  y  =  - ,  u,  v  both  being  functions  of  x. 


Then 


u  -f  Au      u         Au         Av 

! V U 

A?/      v  -\-  Av      v  _     Ax         Ax 
Ax  ~~         Ax  v(y  +  Av) 


Passing  to  the  limit,  we  obtain  the  result 

vdu-udv 

d  (u\-    dx        dx  (5) 

dx\v  J  v1 

TJie  derivative  of  a  fraction,  the  quotient  of  two  functions,  is 
equal  to  the  denominator  multiplied  by  the  derivative  of  the  nu- 
merator minus  the  numerator  multiplied  by  the  derivative  of  the 
denominator,  divided  by  the  square  of  the  denominator. 


DIFFERENTIATION   OF   THE   ELEMENTARY   FORMS      29 

13.   Differentiation  of  a  commensurable  power  of  a  function. 
Let  y  =  un,  in  which  it  is  a  function  of  x.     Then  there  are 
three  cases  to  consider : 

1.  n  a  positive  integer. 

2.  n  a  negative  integer. 

3.  n  a  commensurable  fraction. 

1.  n  a  positive  integer. 

This  is  a  particular  case  of  (4),  the  factors  u,  v,  w,  •••  all 

being  equal.     Thus 

dy  n_xdu 

dx  dx 

2.  n  a  negative  integer. 

Let  n  =  —  m,  in  which  m  is  a  positive  integer. 

Then  y  =  un  =  u~m  =  — , 

«*  dl  =  ^-dl         by  (5),  and  Case  (1) 


hence 


dx 

u2m 

dx 

—  mu~m~ 

idu, 
dx' 

dy  _ 

dx 

n-l  dtt 

wit"     — • 
dx 

3.    7i  a  commensurable  fraction. 

Let  n=*-,  where  p,  g  are  both  integers,  which  may  be  either 

•     q 


positive 

or  nega 

tive. 

p 

Then 

y  =  un  =  u9  ; 

hence 

if  =  fir, 

and 

i.e. 

dec               cte 

30  DIFFERENTIAL   CALCULUS 

Solving  for  the  required  derivative,  we  obtain 

dx     V         dxJ 

hence  -*-Un  =  nun-14".  (6) 

dx,  doc 

The  derivative  of  any  commensurable  power  of  a  function  is 
equal  to  the  exponent  of  the  power  multiplied  by  the  power  with 
its  exponent  diminished  by  unity,  multiplied  by  the  derivative  of 
the  function. 

It  should  be  noticed  that  Vu  =  it2, 

u 

hence  ±^=±*«>        ±(1\  =  =  1*». 

dx  2^/udx         dx\u)       u2  dx 

These  theorems  will  be  found  sufficient  for  the  differentia- 
tion of  any  function  that  involves  only  the  operations  of  addi- 
tion, subtraction,  multiplication,  division,  and  involution  in 
which  the  exponent  is  an  integer  or  commensurable  fraction. 

The  following  examples  will  serve  to  illustrate  the  theo- 
rems, and  will  show  the  combined  application  of  the  general 
forms  (1)  to  (6). 

ILLUSTRATIVE  EXAMPLES 

t  3  x1  -  2     ~    ,  dy 

1.    y  = ;  find  -*-• 

J        x  +  1  dx 

(x  +  l)-f(3  *»-  2)  -  (3x3  -  2)  L  (x-r  1) 

d/  = ~ 7—TO ^ ^^ 

dx  (x  +  l)2 

±  (3^-2)=^  (3^)--f(2)     (by  2) 
dx  dx  dx 

=  6x.     (by  1,6) 

f(si-l)=£?  =  l.     (by  2) 

ax  dx 


DIFFERENTIATION  OF  THE   ELEMENTARY  FORMS      31 

Substitute  these  results  in  the  expression  for  -f- .     Then 

dx 

dy  =  (x  +  1)6  x  -  (3  x2  -  2)  =  3  x*  +  6  x  +  2 
.    <fc  (*+l)2  (x  +  1)2 

2.   u  =  (3s2  +  2)Vl  +  5s2;  find  ^?. 

^  =  (3  s2  +  2)  -  VI  +  5  s2  +  Vl  +  5s2  •  —  (3  s2  +  2) .     (by  3) 
ds  ds  ds 

—  vTToT2  =  —  (1  +  5  s2)* 
</*  ds  . 

=  l(l+5s2)-^(l  +  5s2)      (byG) 
2  as 

5s 


Vl  +  5  s2 

-^(3s2  +  2>=6s.     (by  6) 
ds 

Substitute  these  values  in  the  expression  for  —     Then 

ds 

du  =  5jjS*  +  2)   ,   6gVrTT72  =  45*»+.16» 
rf*         Vl  +  5s2  Vl~+5s2 


3  v=vr+^  +  vr^g;find  $. 

Vi  +  x2  -  vT^2  d* 

First,  as  a  quotient, 

(VT+T2  -  VI  -x*)^(VTTx*  +  VI^2) 


</a; 


(Vl  +  z2-Vl-x2)2 

(Vi  +  x2+ vi  -  x2) — ( vrr^2  -  vi  -  x2) 

, ^  , (by  5) 

(Vl  +  z2-  Vl-z2)2 

—  ( vr+T2  +  vr^)  =  -^  vrr^2  +  —  vn^2.  (by  2) 

dx  dx  dx 

•-f  VIT^  =  #  (1  +  *2)*  =  1  (1  +  *2)_i  -7-  (1  +  *2)-    (by  6) 
dx  dx  2  dx 

lL(l+x*)=2x.    (by  2  and  6) 
dx 


32  DIFFERENTIAL   CALCULUS 

Similarly  for  the  other  terms.     Combining  the  results,  we  have 

dx        x3  {  VI  -  X*J 

Ex.  3  may  also  be  worked  by  first  rationalizing  the  denominator. 


EXERCISES 

Find  the  ^-derivatives  of  the  following  functions : 


1. 

y  =  xxo. 

2. 

y  —  x~s. 

3. 

y  =  cVx. 

4. 

2        Vi 

''       Vx       3 

5. 

y  =  v/^'~5- 

6. 

y  =  (x  +  a)n. 

7. 

y  —  xn  -\-  an. 

8. 

X 

y  =   . 

14. 

y  =  (2  a\  +  x2)  Vai  +  ^i. 

15 

y-i      *     r 

|l  +  Vl-**J 

16. 

1.7- 

:rn  -f  1 

v  —  — - — 

S       xn  -  1 

18 

1                    1 

(a  +  x)m     (b  +  x)w 

19. 

3x3+  2 

v  =  — - 

Va2  -  x*  X(xs  +  !)| 

9    y==«jL?.  20.   y  =  3(x2  +  l)t(4x2-3). 


x2  +  2 


21.    y  =  3  m"  -  7. 


10.  2/=(*  +  l)vW2.  22>   y  =  4w3_6w2+1oM_3. 

11.  Va  +  a  23.   y  =(1-3  u2  +  6u4)(l  +  u2)3. 
v«  +  V*  24.    7/  =  us. 

i2.  y  =  JT±J  25-  $  =■  M*  + 3  *w2  +  *4- 

1~X"  26.    y  =        ""        . 

13.,= *==.  («  +  *>• 

x  +  Vl  —  x2  27.    y  =  t*2j?8io. 

28.  Given    (a  +  a:)5  =  a5  +  5  a4x  +  10  «3-r2  +  10  a2x3  +  5  a*4  +  x* ; 
find  (a  +  x)A  by  differentiation. 

29.  Show  that  the  slope  of  the  tangent  to  the  curve  y  =  x3  is  never 
negative.     Show  where  the  slope  increases  or  decreases. 


DIFFERENTIATION   OF   THE   ELEMENTARY  FORMS      33 

30.  Given  b2x2  +  a2y2  =  d2b2,  find  dJL  :  (1)  by  differentiating  as  to 

dx 

x ;  (2)  by  differentiating  as  to  y ;  (3)  by  solving  for  y  and  differen- 
tiating as  to  x.     Compare  the  results  of  the  three  methods. 

31.  Show  that  form  (1),  p.  25,  is  a  special  case  of  (3),  p.  27. 

32.  At  what  point  of  the  curve  y2  =  ax3  is  the  slope  0  ?  —  1  ?  +  1  ? 

33.  Trace  the  curve  y  =  x3  +  3x2  +  x  —  1. 

34.  y  -  -3^2  +  7     and  u  =  5  x2  -  1 ;  find  (Jl . 

V7  u2  +  5  tf* 

35.  At  what  angle  do  the  curves  y2  =  \2x  and  y2+  x2  +  0  ;r  —  G3  =  0 
intersect  ? 

14.  Differentiation  of  implicit  functions.  When  a  functional 
relation  between  x  and  y  cannot  be  readily  solved  for  y,  the 
preceding  rules  may  be  applied  directly  to  the  implicit  function. 
The  derivative  will  usually  contain  both  x  and  ?/.  Thus  the 
derivative  of  an  algebraic  function,  defined  by  equating  a  poly- 
nomial in  x  and  y  to  zero,  may  be  obtained  by  the  process  illus- 
trated in  the  following  examples  : 

Ex.  1.     Given  the  function  y  of  x,  defined  by  the  equation 

x5  4-  yr°  —  5  xy  +  1  =  0, 


find^. 
dx 

Since 

dx 

-  5  xy  +  1)  = 

0, 

hence 

5x4  +  5</^-5 
.  dx 

y  - 

r      d 
-  5  x  ,-  = 
dx 

o, 

G>: 

f2, 

3) 

Solving  for    -*. ,  we  obtain 
dx 

<]y  _ 

dx 

•'■4 

x  - 

-7/ 

Ex.  2. 

xy2  -\-  x2y  =  1.     Find 

dy 
dx 

Ex.  3. 

*+  y  +  (x-y)2+(2 

x  — 

3  2/)3  =  0. 

Find 

EL.    CALC.  — 3 

34  DIFFERENTIAL   CALCULUS 

15.  Elementary  transcendental  functions.  The  following  func- 
tions are  called  transcendental  functions  : 

Simple  exponential  functions,  consisting  of  a  constant 
number  raised  to  a  power  whose  exponent  is  variable,  as 
4X,  a*  ; 

the  logarithmic  functions,  as  loga  x,  log6  u ; 
the  incommensurable  powers  of  a  variable,  as  x^2,  un ; 
the  trigonometric  functions,  as  sin  u,  cos  u ; 
the  inverse  trigonometric  functions,  as  sin-1  u,  tan-1  x. 

There  are  still  other  transcendental  functions,  but  they  will 
not  be  considered  in  this  book. 

The  next  four  articles  treat  of  the  logarithmic,  the  two  ex- 
ponential functions,  and  the  incommensurable  power. 

16.  Differentiation  of  loga  od  and  loga  u. 


Let 

y  =  loga  x. 

Phen 

y-\-Ay  =  loga  (x  +  Ax), 

Ay      loga  (x  +  Ax)  —  loga  x 

Ax                     Ax 

Liog/*_±M 

iX  \       X       J 


Ax 

For  convenience  writing  h  for  Ax,  and  rearranging,  we  obtain 

Ay 
Ax 


_l.«Wl+») 

x     h         \        xj 

H-;»»K'+;)i- 


-) 


DIFFERENTIATION   OF   THE   ELEMENTARY   FORMS      35 

SB 

f        h\* 
To  evaluate  the  expression  ( 1  +  -  )   when  h  =  0,  expand  it 

by  the  binomial  theorem,  supposing  -  to  be  a  positive  inte- 
ger m. 

The  expansion  may  be  written 

\       mj  m  1-2        m2  1-2-3  m3 

which  can  be  put  in  the  form 

\        mj  1        2        ^1        2  3 

1     2 
Now  as  m  becomes  very  large,  the  terms  — ,  — ,  •••  become 

in    m 

very  small  and  m  increases  without  limit  as  h  approaches  zero. 
As  m  =  cc  the  series  approaches  the  limit 

1  +  1+—  +  —  +  —  +  •••, 

2  !      3 !     4 ! 

which  will  be  discussed  later. 

The  numerical  value  of  this  limit  can  be  readily  calculated 
to  any  desired  approximation.  This  number  is  an  important 
constant,  which  is  denoted  by  the  letter  e,  and  is  equal  to 
2.7182818-..;  thus 

lim    A  +  IV  =  e  =  2.7182818-...* 

*  This  method  of  obtaining  e  is  rather  too  brief  to  be  rigorous ;  it  assumes 

that  —  is  a  positive  integer,  but  that  is  equivalent  to  restricting  Ax  to  ap- 
Ax 

proach  zero  in  a  particular  way.  It  also  applies  the  theorems  of  limits  to  the 
sum  and  product  of  an  infinite  number  of  terms.  The  proof  is  completed  on 
p.  315  of  McMahon  and  Snyder's  "  Differential  Calculus," 


36  DIFFERENTIAL   CALCULUS 

The  number  e  is  known  as  the  natural  or  Naperian  base; 
and  logarithms  to  this  base  are  called  natural  or  Naperian  log- 
arithms. Natural  logarithms  will  be  written  without  a  sub- 
script, as  logcc;  for  other  bases  a  subscript,  as  in  \ogax,  will 
generally  be  used  to  designate  the  base.  The  logarithm  of  e  to 
any  base  a  is  called  the  modulus  of  the  system  whose  base  is  a. 

X 

li      /        7A* 
If  the  value  ,  ™  f  1  +  - )  =  e  is  substituted  in  the  expres- 


sion for  _?,  the  result  is 
dx 


dy     1    , 

/  =  -  •  log,  e. 

ax     x 


More  generally,  by  Art.  8, 


d   ,  logffle  du  fms 

slog„M  =  -s-^.  (7) 

In  the  particular  case  in  which  a  =  e, 

d   .  _  1  du  /ox 

doc  ~~  u  dx 

The  derivative  of  the  logarithm  of  a  function  is  the  product  of 
the  derivative  of  the  function  and  the  modulus  of  the  system  of 
logarithms,  divided  by  the  function. 

17.    Differentiation  of  the  simple  exponential  function. 

Let  /      y  =  au. 

Then  log  y  =  u  log  a. 

Differentiating  both  members  of  this  identity  as  to  x,  we  obtain 
1  dy      ,  du       ,,     oX 

dy      ,  du 

therefore  -j-  <*>u  =  l«g  a  •  a'1 "  ^  ■  ^ 


DIFFERENTIATION   OF  THE   ELEMENTARY   FORMS      37 

In  the  particular  case  in  which  a  =  e, 

d  du  ,  +  n  x 

as  •"=•"•  a*-  (10) 

The  derivative  of  an  exponential  function  with  a  constant  base 

» 
is  equal  to  the  product  of  the  function,  the  natural  logarithm  of 

the  base,  and  the  derivative  of  the  exponent. 

18.    Differentiation  of  an  incommensurable  power. 

Let  y  =  un, 

in  which  n  is  an  incommensurable  constant.     Then 

log  y  =  n  log  u, 

1  dy  _  n  du 

ydx      u  dx' 

dy_  y     du 

dx  u     dx' 

d     _  __,  du 
dx  dx 

This  has  the  same  form  as  (6),  so  that  the  qualifying  word 
"  commensurable  "  of  Art.  13  can  now  be  omitted. 


EXERCISES 
Find  the  x  derivatives  of  the  following  functions 


1. 

y- 

=  log(x+  a). 

2. 

y- 

=  log  (ax  +  b). 

3. 

y  = 

=  log  (4  x2  -  7  x 

4-2) 

4. 

y  = 

=  logj  +  *. 

1  —  X 

5. 

2/  = 

1  —  X2 

6. 

y  = 

=  x  log  X. 

7. 

y  = 

-  xn  log  X. 

8. 

y  = 

-  xn  log  xm.    N 

9.    y  =  log  Vl  —  X2. 
10.    y  =  Vx  —  log  (  y/x  +  1) . 


11.  y  =  \oga(3x2-V2  +  x). 

12.  y  =  log10  (x2  +  7  x). 

13.  y  =  logx  a. 

14.  y  =  exa. 

154  y  =  e4*+5. 

l 
16.   y=-el+x. 


38 


DIFFERENTIAL   CALCULUS 


17.  y  =  — - — . 

18.  y  =  ex(l  -z«). 

-X  .  —  X 

19.  y 


23. 

24. 


log 


Vq  -f  Vx 


e*  +  e~x 

20.  y  =  log  (e*  -  c-*). 

21.  y  =  log  (x  +  ex). 

22.  v  =  xna*. 


Va 
1 
logx' 

25.  3/=  (log*)*. 

26.  y  =  log  (log a). 

27.  y 

28.  y  =  al°s 


x  log  - . 
x 


The  following  functions  can  be  easily  differentiated  by  first  taking 
the  logarithms  of  both  members  of  the  equations. 

31.  ¥  =*&+*!. 

32.  y  =  x\a  -f  3  x)\a  -  2  x)2. 

33.  y  =  V(*+g)'. 

Vx  —  a 


29. 


(*-*)■ 


(a;-2)*(a;-3)* 


30.   y  =  xVl  -  x(l  +  a:). 


19.    Limit  of 


sin 


0 


as  9  approaches  0.     Before  proceeding  to 


determine  the  derivatives  of  the  trigonometric  functions  it  is 
necessary  to  prove  the  following  lemma  : 
lim   sin  0 


0 


=  1. 


0  =  0 

With  0  as  a  center  and  OA  =  r 

as   radius,   describe    the    circular 

arc   AB.     Let  the  tangent  at  A 

-4  meet   OB  produced  in  D;    draw 

Fig.  7  BC  perpendicular  to  OA,  cutting 

OA  in  C.     Let  the  angle  OAB  =  0  in  radian  measure, 

then  arc  AB  =  r#, 

(75  <  arc  AB  <  .4Z),  by  geometry 

i.e.  r  sin  0  <  r6  <  r  tan  0, 

.  sin  6  <  0  <  tan  0. 


DIFFERENTIATION   OF   THE   ELEMENTARY   FORMS      39 

By  dividing  each  member  of  these  inequalities  by  sin  0, 

1  <^-<sec0; 
sin  0 


but  sec  0  =  1,  when  0  =  0 

0 
sin  0 


,  lim       0         .,         A    lim   sin  0      -, 

hence>  0=o^  =  1> and  »=o-vr  =  1 


20.   Differentiation  of  sin  ?/. 
Let  y  =  sin  u. 

,p.  Ay  _  sin  (u  4-  A?/)  —  sin  w     Ait 

A./1  Aw  A/ 

To  evaluate  the  expression 

sin  (u  +  Aw)  —  sin  w, 

we  make  use  of  the  formulas  for  the  sine  of  the  sum  and  the 
sine  of  the  difference  of  two  angles.     Since 

sin  (a  +  b)  =  sin  a  cos  b  4-  cos  a  sin  b, 
sin  (a  —  b)  =  sin  a  cos  6  —  cos  a  sin  6, 

hence,  by  subtracting  the  second  equation  from  the  first, 
sin  (a  +  b)  —  sin  (a  —  b)  =  2  cos  a  sin  b. 

This  equation  is  true  for  all  values  of  a  and  of  b.    In  particu- 
lar, then,  putting 

'  '  F         5  a  +  6  =  w  +  Aw, 

and  a  —  b  =  u, 

that  is,  a  =  u  +  — ,  and  6  =  — , 

we  obtain 

sin  (w  +  Aw)  —  sin  w  =  2  cos  (u-\ — -  )  sin  — . 


40 


DIFFERENTIAL   CALCULUS 


The  expression  for  — J.  may  now  be  written  in  the  form 

Ax 

Aw 
sm  — 
A?/  /         Au\  2   A?* 

— ^  =  cos  [u  +  — — - , 

Ax  V  2  J       Au     Ax 


hence 


dy  .. 

-f-  =  cos  u  •  li m 

dx  Aw  =  0 


sm 


2 
Aw' 


Aw 
2~ 


c?w 

dx~' 


hence,  by  Art.  19,      ^-  sin  w  =  cos  u  — 
rfx  doc 


(ID 


77>e  derivative  of  the  sine  of  a  function  is  equal  to  the  product 
of  the  cosine  of  the  function  and  the  derivative  of  the  function. 

21.   Differentiation  of  cos  u. 

Let  ?/  =  cos  u  —  sinf  -  —  u  \ 

bl 

dx     dx 

du 


-w)-^(     -u\ 


dx\2 


d 

—  i 

doc 


u  =  —  sinu 


doc 


(12) 


The  derivative  of  the  cosine  of  a  function  is  equal  to  minus  the 
product  of  the  sine  of  the  function  and  the  derivative  of  the  function. 

22.   Differentiation  of  tan  u. 


Let 


Then 


y  =  tan  u 


sin?* 
cos?* 


d    .  d 

cos  u  •  —  sin  u  —  sin  u  •  — cos  u 


dy                 dx 

dx 

-     (bv  5) 

dx  ~~                       cos-  u 

•    o      du  .     •   Q  .   du 

cosJ  u h  sm-  U  — 

dx                 dx 

du 
dx 

(by  11,  12) 

cos2  u 

cos2  u 

that  is, 


^tant*  =  sec2t*^ 
doc  doc 


(13) 


DIFFERENTIATION   OF  THE   ELEMENTARY   FORMS      41 

The  derivative  of  the  tangent  of  a  function  is  equal  to  the 
product  of  the  square  of  the  secant  of  the  function  and  the  deriva- 
tive of  the  function. 

Since  the  remaining  elementary  trigonometric  functions 
can  be  expressed  as  rational  functions  of  those  already  con- 
sidered, their  derivatives  can  be  obtained  by  means  of  the 
preceding  rules.     The  results  are 

4-wtu=  -csc2t*^.  (14) 

dx  dx 

~  sec  u  =  sec  u  tan  u  —  •  (15) 

dx  dx 

lL  esc  u  =  -  esc  w  cot  w  —  (16) 

dx  dx 

EXERCISES 

Find  the  ^-derivatives  of  the  following  functions : 

1.  y  —  sin  7  x. 

2.  y  —  cos  5  x 

3.  y  =  sin  x2. 

4.  y  =  sin  2  x  cos  x. 

5.  y  =  sin8  x. 

6.  y  =  sin  5  x2. 

7.  y  =  sin2  7  x. 

8.  y  —  \  tan3  x  —  tan  x. 

9.  y  =  sin3  x  cos  x. 

10.  ?/  =  tan  x  +  sec  x. 

11.  y  =  sin2  (1-2  x2)2. 

12.  #  =  tan  (3  -5  a;2)2. 

13.  y  —  tan2  a:  —  log  (sec2  a;). 

14.  y  =log  tan  ($#  +  Jir). 
■  15.   y  =  logsinVx.  29.    ?/  —  tan  (x  -f  y) . 


16. 

*    i 
y  =  tan  a*. 

17. 

y  =  sin  narsinna;. 

18. 

y  =  sin  (w  +  b)  cos  (w  —  J). 

19. 

sinm  nx 

y  = . 

cosnmx 

20. 

y  =  x  +  log  cos  (  x  -  j  J. 

21. 

?/  =  sin  (sin  t/). 

22. 

?/  =  sin2^. 

23. 

y  =  sin  ex  •  log  a\ 

24. 

?/  =  Vsin  x2. 

25. 

y  =  esc2  4  x. 

26. 

y  —  sec(4  x  —  3)2. 

27. 

r/  =  cot  x2  -f  sec  Var. 

28. 

y  ~  sin  a:?/. 

42  DIFFERENTIAL   CALCULUS 

30.  Find  —  (cos  u)  directly  from  the  definition  of  the  derivative. 

dx 

Also  — (tan  u). 
dx 

31.  Find  —  (cos  u)  from  the  relation  sin2  u  +  cos2  u  =  1. 

dx 

23.    Differentiation  of  sin-1  u- 

Let  y  =  sin~1u. 

Then  sin  y  =  u, 

and,  by  differentiating  both  members  of  this  identity, 

dy    ,du 

cos  y  —  =  — ; 

d#     dx 


hence 


dy  _     1     fZw 1  d?t. 

da;      cos  ?/  d.r      _j_  Vl—  sin2y  dx ' 

d    •   _i  1        dw 


i.e.  — -  sm-1  it  =  ± 

d#  VI  _  tf  d^ 

The  ambiguity  of  sign  accords  with  the  fact  that  sin-1  u  is  a 
many- valued  function  of  u,  since,  for  any  value  of  u  between 
—  1  and  1,  there  is  a»series  of  angles  whose  sine  is  u :  and,  when  u 
receives  an  increase,  some  of  these  angles  increase  and  some 

decrease;  hence,  for  some  of  them,       sm — -  is  positive,  and 

du 

for  some  negative.  It  will  be  seen  that,  when  sin-1  u  lies  in 
the  first  or  fourth  quarter,  it  increases  with  u,  and,  when  in 
the  second  or  third,  it  decreases  as  u  increases.  Hence,  for  the 
angles  of  the  first  and  fourth  quarters, 

jL  Sin-1!*  =  --f- COS"1  !*:=  +  ■— 1—  gg.  (17) 

€loc  doc  VI  —  «*2  dor 

In  the  other  quarters  the  minus  sign  is  to  be  used  before 
the  radical. 


DIFFERENTIATION   OF   THE   ELEMENTARY  FORMS      43 

The  derivatives  of  the  other  inverse  trigonometric  functions 
can  be  easily  obtained  by  the  method  employed  in  the  present 
article.  The  most  important  of  the  remaining  ones  are  tan-1  u, 
sec-1 u ; 

4-  tan"1  u  =  -  4-  cot1  u  =  -L-^. 


doc  ddc  l  +  u2  doc 


x  sin-1  a;. 


2.  y  =  cos-1Vl  —  x'2.  17#   y 

3.  y  =  sin-i  (3  x  -  1).  18.   y  =  e**"1*. 

4.  y  =  sin-1  (3x  -4  x3).  1Q _.,        1 

1  —  x2 

5.  y  =  sm-i- — -• 

1  +  r 


6.  y  =  vsin-1  x. 

1.  y  —  tan-1  ?*. 

8.  y  =  cos-1  log  x. 

9.  y  =  sin-1  (tan  a-). 
10.  y  =  sec-1  — 


VI  -x2 

11.  #  =  CSC-1  -• 

a: 

12.  y^tan^f        1 

V  Va:2  -  1 

13.  y  =  tan-1-* 

14.  y  =  sin"1  Vsin  x. 

15.  y  =  tan-1V^^ 

1  -f  cos 


(18) 


-^-sec  ~1u  =  --^-csc-1m  = !__*•.  (19) 

das  dx  u^u2—l(lQC 


EXERCISES 

Find  the  ar-derivatives  of  each  of  the  following  functions : 
1.   y  =  sin-1  2  x2.  16.    y  =  tan  a;  •  tan-1  x. 


2  a;2  -  1 

20. 

y  =  sec"1  *L±i. 
^                a;2-  1 

21. 

y-tan-lV^+  V«. 
1  —  V«ar 

22. 

y  =  cos  x 

23. 

y  =  tan-1  (n  tan  a). 

24. 

y  =  cos-1  (cos  2  a-). 

25. 

y  =  cos-1  (2  cos  a:). 

26. 

y  =  tan-1  (Vl  +  x2  - 

X). 

27. 

y  =  2  tan"1  aL— ' 
M  +  a: 

28. 

y  =  tan-1        — |- tan 
iV3 

.^b-x 
a:V3 

44  DIFFERENTIAL   CALCULUS 

24.   Table  of  fundamental  forms. 

d(cu)  =cdu.  (1) 

dx  dx 

Jt{u  +  V-.w)=du+dv_dw.  2 

dx  dx     dx      dx 

*&»1  =u^  +  v^  (3) 

dx  dx        dx 

d-(uvw)  =uvm  +  vwm  +  wu<M.  (4) 

dx  dx  dx  dx 


vdu_udv 
d  u  _     dx         dx 

dx  v  »,2 


(5) 


^aw  =loga-au    ^  (9) 

da?  da? 


Ae«  =eud^                                                (10) 

<ia?  da? 

-f-sinw  =costi~>                                           (11) 

<?a?  da? 

^-cosi*  =-smudw.                                       (12) 

c?a?  da? 

^tanw  =sec*tt^.                                         (13) 

da?  da? 

-£-  cot  ft  =  -  esc2  u  d^                                     (14) 

da?  da? 


DIFFERENTIATION   OF   THE   ELEMENTARY   FORMS      45 


4-  sec  u      =  sec  u  tan  u  ^-  (15) 

doc  doc 

^cscu      =  -cscMcotw~  (16) 

doc  doc 

JL  sinl  u=-4-  COS"1  M  = 1 *L  (17) 

«a?  «a?  Vi  _  ^2 «a? 

4-  tan"1 t*  =  -  #  cot1  w  =  —1—  *!.  (18) 

dx  doc  i  +  w2  da? 

4-  sec1  u=-~-  esc1 t*  = -J ^*  (19) 

doc  doc  uy/u*  -id& 


EXERCISES  ON   CHAPTER   II 

Find  the  ^-derivatives  of  the  following  functions  : 
1.   y  =  3  Z2  +  5  a,-3  -  7. 
o  3       5      1 


3.    y  =  (x  +  5)  Vx  -  3. 


4.  ?/  =  xVa'2  —  x2. 

5.  ?/  =  x  log  sinx. 


a 


10. 

y  =  log--       . 
a* 

11. 

1-x2 

y  = 

Vl  +  x2 

12. 

y  =  ex  cos  x. 

13. 

,  =  00^(1). 

14. 

.       ,      4  sin  x 
?/  =  tan-1 

3  +  5  cos  x 

15 

v  —  (x  4-  rtH.an-1-!/- 

6-     y=-V«2_x2. 
(  r2    ? 

>  7.    y=c- ec- 
x 

8.   J,=  ta„2z,z  =  t.n-i(2»-l).  ^ 

.  16.    w  =  cot  x — — — 

2    9.   y  =  eVu,  u  =  log  sin  x.  a; 

17.  ?/  =  tan4  x  —  2  tan2  x  +.log(sec*  x). 

18.  y  =  £l2«£+log(l-a;). 


19.    y  =  C0S-i3  +  5c0S:C. 
5  +  3  cos  x 


46  DIFFERENTIAL   CALCULUS 

20.  y.=  log  (l±£^-|tan-i*. 

21.  #  =  log  (x  +  Vx2  -  a2)  +  sec"1-- 

a 

22.  y  =  eu,  u  —  log  x.  25.   x2?/2  +  x8  +  y3  =  0. 

23.  y  =  log  s2  +  e",  s  =  sec  x.  26.   x3  -f  x  =  y  +  y8. 

24.  xs  +  ys  —  3  axy  =  0.  27.   xy2  +  x2#  —  x-\-y. 

28.   y  =  sin  (2  u  -  7),  u  =  log  x2. 

29.  By  means  of  differentiation  eliminate  the  constant  p  from  the 
equation  y  =px2. 

30.  At  what  points  is  the  tangent  to  the  curve  y  =  cos  x  parallel  to 
the  x-axis  ? 

31.  Show  that  the  x-derivative  of  tan-1  \    ~~  cos  x  is  not  a  func- 

*  1  +  cos  X 
tion  of  x. 

x2      ?/2 

32.  Find  at  what  points  of  the  ellipse h  *—  —  1  the  tangents  cut 

_,  .  a2      Z>2 

off  equal  intercepts  on  the  axes. 

33.  Find  the  points  at  which  the  slope  of  the  curve  y  =  tan  x  is 
twice  that  of  the  line  y  =  x. 

34.  Find  the  angle  which  the  curves  y  =  sin  x  and  y  =  cos  x  make 
with  each  other  at  their  point  of  intersection. 


CHAPTER   III 

SUCCESSIVE  DIFFERENTIATION 

25.  Definition  of  nth  derivative.  When  a  given  function 
y~<f>(x)  is  differentiated  with  regard  to  x  by  the  rules  of 
Chapter  I,  then  the  result    dy 

is  a  new  function  of  x  which  may  itself  be  differentiated  by  the 
same  rules.     Thus,         7  , ,  .         , 


dx\dxj     dx 

The  left-hand  member  is  usually  abbreviated  to  _J?,  and  the 

dx2 
right-hand  member  to  <f>"(x) ;  that  is, 

Differentiating  again  and  using  a  similar  notation,  we  obtain 
±(cPy\_d*i_ 


dx\dx2J     dx3 

and  so  on  for  any  number  of  differentiations.     Thus  the  sym- 

d2y 

bol  -r^  expresses  that  y  is  to  be  differentiated  with  regard  to  x, 

and  that  the  resulting  derivative  is  then  to  be  differentiated. 

dsy   .      .  d 

Similarly,  -^  indicates  the  performance  of  the  operation  — 

three  times,  j-(  j~(;p  ))•     ^n  general,  the  symbol  -~  means 

that  y  is  to  be  differentiated  n  times  in  succession  with  regard 
to  x. 

47 


48  DIFFERENTIAL   CALCULUS 

Ex.  1.     If  y  =  x4  +  sin  2  x, 

-£  =  4  x8  +  2  cos  2  x, 
ufx 

g  =  12x2-4sin2x, 

?^  =  24x-8cos2x, 
«x3 

^  =  24  + 16sin2x.  ■ 

dx4 

If  an  implicit  equation  between  x  and  y  is  given  and  the 
derivatives  of  y  with  regard  to  x  are  required,  it  is  not  neces- 
sary to  solve  the  equation  for  either  variable  before  perform- 
ing the  differentiation. 

Ex.  2.     Given  x4  +  y4  +  4  a2xy  =  0  ;  find  v^ . 

dx'2 

^-(x*  +  y*+±a2xy)=0, 

4  x8  +  4  ?/3  ^  +  4  a2*  ^  +  4  a2#  =  0. 

The  last  equation  is  now  to  be  solved  for  -j-, 
dy         xs  +  a2y 


dx  ys  +  a2x 

Differentiating  again,  we  obtain 


(i) 


d2y  _       d  fx8  +  a2y\ 
dx2~      dx\y8  +  a2x] 


w 

(y8  +  a2x)  —  (x3  +  a2y)  -  (x8  +  a*y)  JL  (y8  +  a2x) 
dx  '     dx 

(yTT~a27y2 

(y8  +  a2x)  ('dx2  +  a2'-1!')  -  (x8  +  a2?/)  fsv2^  +  a") 
V  '/x/  "      V         ax         / 

(*/3  +  d2x)2 


SUCCESSIVE   DIFFERENTIATION  49 

The  value  of  JL   from  (1)  is  now  to  be  substituted  in  the  last 
dx 

equation,  and  the  resulting  expression  simplified.     The   final   form 
may  be  written  : 

dhj      2  a*xy  -  1 0  a2xsys  -  a4  O4  +  y4)  -  3  x2y2  O4  +  #4) 
dx2  ~  (y8  +  a2x)8 

In  like  manner  higher  derivatives  may  be  found. 

26.  Expression  for  the  nth  derivative  in  certain  cases.  For  cer- 
tain functions,  a  general  expression  for  the  nth  derivative  can 
be  readily  obtained  in  terms  of  n. 

where  n  is  any  positive  integer.     If  y  =  eax,  ~-  =  aneax. 


Ex.  2. 

If 

y  '=  sin  x, 

dy                        1        tt\ 
^|=cos:r  =  sin^+^j, 

g=cos(x  +  |)=sin(x+^), 
6/nv        .     /         n7r\ 

If 

y~- 

-  sin  ax. 

dny                 1           mr\ 
d^  =  aSln[ax+2-)- 

EXERCISES  ON  CHAPTER   III 

1.  y=3x*+5x2+3x-9;  find^.  5.  ij  =  tana:;  find  ^l. 

dx3  J  dx8 

2.  y  =  2x2  +  Zx  +  5;  find  v|.  6.  y  =  e*logz;  find  ^X 

3'  ^  =  *;  fmdS'  7."y  =  *«loga:j  find  g. 

4.  y  =  ^  _I2;  find  g.  8.  y  =  sec**;  find  g. 

EL.    CALC.  — 4 


50  DIFFERENTIAL  CALCULUS 

9.  y  =  logsina:;  find  g.  18.  3^^-p  find  J*. 

10.  y  =  sin  x  cos  a; ;  find  -r\>  19.  y  z=  cos  ma: ;  find  ~^- 

U.  ,  =  ^L;  flnd  %  20.  y  =  ,      X    .    ;  find  £». 

12.  y  =  **  logs';  find  ^.  21.  y  =  log  (a  +  *)";  find  J.    , 

13.  y  =  sin,;  find  g.  22.  y2  =  2^;  find  g- 

14.  y  =  log  (*  +  *-)  ;  find  g  •       23.   J  +  £  =  1 ;  find  g. 

15.  y  =  (^- 3^  +  3)^;  find  g.    24.  a*  +  y3  =  3axy;  find  g-      t 

16.  3/  =  a;4  log  a;;  find  g.  25.  «■+»  =  zy;  find  g. 

17.  y  =  eax  I  find  g-  26.  y  =  1  +  xe*  ;  find  ^- 

d  u  (in 

27.  y  =  «*sina;;  prove  ^J,-2-^+2y  =  0. 

28.  y  =  aa;sina;;  prove  x2~  -  2a;  -£  +  02  +  2)y  =  0. 

d2y 

29.  y  =  axn+1+  bx~n\  prove  a;2^J>  =  n(n  +  l)y. 

30.  y  =(sin-^)2;  prove  (1  -  z2)  ^2  -  X-+  =  2. 

»■»  =  ££  P^g-1-* 

_    ,  dny 

32.  y  =  xn_1loga:;  find  -=-£• 

33.  „  =  *.;  prove  ^  =  2^-^  +  9«-. 

34.  y  =  cos2 a:;  find  -^. 


CHAPTER   IV 

MAXIMA   AND    MINIMA 

27.  Increasing  and  decreasing  functions.  A  function  is  said 
to  be  increasing  if  it  increases  as  the  variable  increases  and 
decreases  as  the  variable  decreases.  A  function  is  said  to  be 
decreasing  if  it  decreases  as  the  variable  increases  and  increases 
as  the  variable  decreases.  When  the  graph  of  the  function  is 
known  it  will  indicate  whether  the  function  is  increasing  or 
decreasing  for  an  assigned  value  of  x ;  conversely,  a  knowledge 
of  the  fact  whether  a  function  is  increasing  or  decreasing  is  of 
great  assistance  in  drawing  the  graph.  Usually  a  function  is 
increasing  for  certain  values  of  x  and  decreasing  for  others. 

28.  Test  for  determining  intervals  of  increasing  and  decreasing. 

Let  y  —  4>(x)  be  a  continuous  function  having  a  derivative  for 
all  values  of  x  from  a  to  b.  By  the  above  definition  y  is  in- 
creasing or  decreasing  at  a  point  xl,  according  as 

k  =  Qfa  +  h)  -  <£(zi) 

has  or  has  not  the  same  sign  as  h,  where  h  is  a  sufficiently 
small  number.  Hence  <f>(x)  is  an  increasing  or  a  decreasing 
function  at  the  value  xx  according  as 

is  positive  or  negative. 

51 


52 


DIFFERENTIAL   CALCULUS 


Thus,  the  function  y  =  <f>(x)  is  increasing,  if  <f>'(x)  is  positive ; 
if  4>'(x)  is  negative,  the  function  is  decreasing. 

In  order  that  a  function  shall  change  from  an  increasing 
function  to  a  decreasing  function  or  vice  versa,  it  is  necessary 
and  sufficient  that  its  derivative  shall  change  sign.  If  the 
derivative  is  continuous,  this  can  happen  only  when  the  deriva- 
tive passes  through  the  value  zero.  The  derivative  may  also 
change  sign  when  it  becomes  infinite,  and,  notwithstanding 
this  discontinuity  of  the  derivative,  the  original  function  may 
still  be  continuous.  In  the  graph  of  the  function  this  requires 
that  at  such  a  point  the  tangent  to  the  locus  shall  be  parallel  to 
the  y-axis.     The  process  will  be  illustrated  by  a  few  examples. 

Ex.   Find  the  intervals  in  which  the  function 
<f>(x)  =  2  a;8-  9a;2  +  12a;  -  6 
is  increasing  or  decreasing.     The  derivative  is 

<f>'(x)  =  6a;2  -  18 a;  +  12  =  Q(x  -  \){x  -  2); 

hence,  as  x  passes  from  -co  to  1,  the  derived  function  <f>'(x)  is  posi- 
tive and  <f>(x)  increases  from  <£(  — oo) 
to  <£(1),  i>e.  from  <f>=—ccto<f>=  — 1; 
as  x  passes  from  1  to  2,  <f>'(x)  is  nega- 
tive, and  <f>(x)  decreases  from  <£(1)  to 
<£(2),  i.e.  from  —  1  to  -  2;  and  as  x 
passes  from  2  to  +  co,  <£'(*)  is  posi- 
tive, and  <f>(x)  increases  from  <£(2)  to 
<£(ao),  i.e.  from  —  2  to  +  oo .  The 
locus  of  the  equation  y  =  <£(a;)  is  shown 
in  Fig.  8.  At  points  where  <f>'(x)  =  0, 
the  function  <f>(x)  is  neither  increas- 
ing   nor   decreasing.     At   such   points 

the  tangent  is  parallel  to  the  axis  of  x.     Thus  in  this  illustration,  at 

x  =  1,  x  =  2,  the  tangent  is  parallel  to  the  x-axis. 


Fig.  8 


MAXIMA   AND   MINIMA  53 

EXERCISES 

1.  Find  the  intervals  of  increasing  and  decreasing  for  the  function 

<f>(x)  =  xs  +  2x2  +  x  -  4. 
Here  <£'(*)  =  3x2+4.r+l  =  (3z+  l)(x  +  1). 

The   function   increases  from   x  =  —  coto  x  =  —  1 ;   decreases  from 
a:  =  —  ltoar=-—  |;  increases  from  a:  =  —  \  to  a:  =  go  . 

2.  Find  the  intervals  of  increasing  and  decreasing  for  the  function 

y  =  Xs  -  2  x2  +  x  -  4, 
and  show  where  the  curve  is  parallel  to  the  a>axis. 

3.  At  how  many  points  can  the  slope  of  the  tangent  to  the  curve 

#  =  2x3-3x2  +  l 
be  1  ?  - 1  ?     Find  the  points. 

4.  Compute  the  angle  at  which  the  following  curves  intersect  ; 

y  =  3  x2  -  1,  y  =  2  x2  +  3. 

29.  Turning  values  of  a  function.  It  follows  that  the  values 
of  x  at  which  <f>  (x)  ceases  to  increase  and  begins  to  decrease 
are  those  at  which  <j>'(x)  changes  sign  from  positive  to  nega- 
tive ;  and  that  the  values  of  x  at  which  <f>  (x)  ceases  to  decrease 
and  begins  to  increase  are  those  at  which  <f>'(x)  changes  its 
sign  from  negative  to  positive.  In  the  former  case,  <£(#)  is 
said  to  pass  through  a  maximum,  in  the  latter,  a  minimum, 
value. 

Ex.  1.     Find  the  turning  values  of  the  function 
<£0)  =  2  xs  -  3  x2  -  12  x  +  4, 

and  exhibit  the  mode  of  variation  of  the  function  by  sketching  the 

curve  ,  ,  x 

V  =  <KX)- 

Here  <f>'(x)=6x2  -  Qx-  12  =  6(z  +  1)0  -  2), 


54 


DIFFERENTIAL   CALCULUS 


Fig.  9 


9,   -  16,   -  5,  36,  oo. 


hence  <j>'(x)  is  negative  when  x  lies  between  -  1  and 
+  2,  and  positive  for  all  other  values  of  x.  Thus  <f)(x) 
increases  from  x  =  -  oo  to  x  =  -  1 ;  decreases  from 
{  x  =—  1  to  a:  =  2;  and  increases  from  x  =  2  to  x  =  oo. 
Hence  </>(-  1)  is  a  maximum  value  of  <f>(x),  and 
<f>(2)  a  minimum. 

The  general  form  of  the  curve  y  =  <f>(x)  (Fig.  9) 
may  be  inferred  from  the  last  statement,  and  from 
the  following  simultaneous  values  of  x  and  y  : 

x  =  -  oo,   -  2,   -  1,  0,       1,         2,       3,     4,  oo. 

y  =  -co,       0,      11,  4, 

Ex.  2.    Exhibit  the  variation  of  the 
function    <f>(x)  =  (x—l)~$+ 2, 
especially  its  turning  values. 

Since  <f>'(x)  =  - -, 

6  (x  -  1)* 

hence    <f>'(x)    changes   sign   at  x  =  l, 

being    negative   when   x  <  1,   infinite 

if  x  =  1,  and  positive  if  z>  1.     Thus 

<f>(l)  =  2  is  a  minimum  turning  value 

of  tf>(x).    The  graph  of  the  function 

is  as  shown  in  Fig.  10,  with  a  vertical  tangent  at  the  point  (1,  2). 

Ex.  3.     Examine  for  maxima  and  minima  the  function 


Fig.  10 


<f>(x)  =  (x 
1 


iy  + 1, 


Here        <f>'(x)  = 


30-l)§ 


Fie.  11 


hence  cf>'(x)  never  changes  sign,  but  is  always 
positive.     There   is  accordingly   no  turning- 
value.    The  curve  y  ='  <f>(x)  has  a  verti- 
cal tangent  at  the  point  (1,  1),  since  -^ 

dx 

is  infinite  when  x  =  l.  (Fig.  11.) 


MAXIMA   AND   MINIMA  55 

30.  Critical  values  of  the  variable.  It  has  been  shown  that 
the  necessary  and  sufficient  condition  for  a  turning  value  of 
<f>(x)  is  that  <f>'(x)  shall  change  its  sign.  Now  a  function 
can  change  its  sign  only  when  it  passes  through  zero,  as  in 
Ex.  1  (Art.  29),  or  when  its  reciprocal  passes  through  zero, 
as  in  Ex.  2.  In  the  latter  case  it  is  usual  to  say  that  the 
function  passes  through  infinity.  It  is  not  true,  conversely, 
that  a  function  always  changes  its  sign  in  passing  through 
zero  or  infinity,  e.g.  x2  and  x~2. 

Nevertheless  all  the  values  of  x,  at  which  <j>'(x)  passes 
through  zero  or  infinity,  are  called  critical  values  of  x,  be- 
cause they  are  to  be  further  examined  to  determine  whether 
<f>'(x)  actually  changes  sign  as  x  passes  through  each  such 
value  ;  and  whether,  in  consequence,  <f>  (x)  passes  through  a 
turning  value. 

For  instance,  in  Ex.  1,  the  derivative  <f)'(x)  vanishes  when 
x  =  —  1,  and  when  x  =  2,  and  it  does  not  become  infinite  for 
any  finite  value  of  x.  Thus  the  critical  values  are  —  1,  2, 
both  of  which  give  turning  values  to  <f>  (x).  Again,  in 
Exs.  2,  3,  the  critical  value  is  x  =  1,  since  it  makes  <f>'(x) 
infinite  ;  it  gives  a  turning  value  to  <£  (x)  in  Ex.  2,  but  not 
in  Ex.  3. 

31.  Method  of  determining  whether  <f>'(&)  changes  its  sign  in 
passing  through  zero  or  infinity.  Let  a  be  a  critical  value  of  x; 
in  other  words,  let  <£'(a)  be  either  zero  or  infinite,  and  let  h 
be  a  very  small  positive  number,  so  that  a  —  h  and  a  +  h  are 
two  numbers  very  close  to  a,  and  on  opposite  sides  of  it.  In 
order  to  determine  whether  <f>'(x)  changes  sign  as  x  increases 
through  the  value  a,  it  is  necessary  only  to  compare  the  signs 
of  <f>'(a  -f-  h)  and  <f>'(a  —  h).     If  it   is   possible   to   take  h  so 


56 


DIFFERENTIAL   CALCULUS 


small  that  cf>'(a  —h)  is  positive  and  <f>'(a  +  h)  negative,  then 
cf>'(x)  changes  sign  as  x  passes  through  the  value  a,  and 
<f>(x)  passes  through  a  maximum  value  <f>(a).  Similarly,  if 
<j>\a  —  h)  is  negative  and  <f>'(a  +  h)  positive,  then  <f>(x)  passes 
through  a  minimum  value  <£  (a). 

If   <f>'(a  —  h)  and   <j>'(a  +  h)  have   the  same   sign,  however 
small  h  may  be,  then  <f>  (a)  is  not  a  turning  value  of  <£  (x). 

Ex.     Find  the  turning  values  of  the  function 

4>0)  =  0-i)20  +  i)8. 

Here  <f>'(x)  =  2(x-  1)0  +  1)8+  S(x  -  l)2(x  +  l)2 

=  (a:-l)(x+  1)2(5  x-  1). 

Hence  <j>'(x)  becomes  zero  at  x  =  —  1,  \,  and  1 ;  it  does  not  become 
infinite  for  any  finite  value  of  x. 

Thus,  the  critical  values  are  —  1,  |,  1. 


Fia.  12 


When  x  has  any  value  less  than   —  1,  the  three  factors  of  <f>'(x) 
take  the  signs  —  +  — ,  hence  <f>'(x)  is  +,  and  when  x  has  a  value 


MAXIMA   AND   MINIMA  57 

between  —  1  and  .J  they  become  —  +  — ,  and  <f>'(x)  is  still  +  ;  hence 
<f>(—  1)  =  0  is  not  a  turning  value  of  <f>(x). 

When  x  has  any  value  between  \  and  1,  the  signs  are  —  +  +  and 
<f>'(x)  is  —  ;  hence  <f>($)  is  a  maximum. 

Finally,  if  x  has  any  value  greater  than  1,  the  signs  are  +  +  +; 
hence  <f>'(x)  changes  sign  from  —  to  +  as  x  increases  through  1,  and 
<£(1)  =  0  is  a  minimum  value  of  <j>(x). 

The  general  march  of  the  function  may  be  exhibited  graphically 
by  tracing  the  curve  y  =  <{>(x)  (Fig.  12),  using  the  foregoing  results 
and  observing  the  following  simultaneous  values  of  x  and  y : 

x^-co,   -  2,   -  1,  0,  \,  1,     2,  go. 

y  =  -  oo,    -  9,        0,  1,  1  •  1  ••-,  0,  27,  oo. 

32.  Second  method  of  determining  whether  <!>'(»)  changes  sign  in 
passing  through  zero.  The  following  method  may  be  employed 
when  the  function  and  its  derivatives  are  continuous  in  the 
vicinity  of  the  critical  value  x  =  a. 

Suppose,  when  x  increases  through  the  value  a,  that  <j>'(x) 
changes  sign  from  positive  through  zero  to  negative.  Its 
change  from  positive  to  zero  is  a  decrease,  and  so  is  the  change 
from  zero  to  negative ;  thus  <f>'(x)  is  a  decreasing  function  at 
x  =  a,  and  hence  its  derivative  <£%e)  is  negative  at  x=  a. 

On  the  other  hand,  if  <f>'(x)  changes  sign  from  negative 
through  zero  to  positive,  it  is  an  increasing  function  and  <f>"(x) 
is  positive  at  x  =  a ;  hence  : 

TJie  function  cf>(x)  has  a  maximum  value  <f>(a),  when  <f>'(a)  =  0 
and  <f>"(a)  is  negative;  $(x)  has  a  minimum  value  <f>(a),  when 
cf>'(a)  =  0  and  <f>"(a)  is  positive. 

It  may  happen,  however,  that  <f>"(a)  is  also  zero. 
.  To  determine  in  this  case  whether  <£(#)  has  a  turning  value, 
it  is  necessary  to  proceed  to  the  higher  derivatives.     If  <f>(x)  is 

4 


58  DIFFERENTIAL   CALCULUS 

a  maximum,  <j>"(x)  is  negative  just  before  vanishing,  and 
negative  just  after,  for  the  reason  given  above ;  but  the  change 
from  negative  to  zero  is  an  increase,  and  the  change  from  zero 
to  negative  is  a  decrease ;  thus  <f>"(x)  changes  from  increasing 
to  decreasing  as  x  passes  through  a.  Hence  (f>'"(x)  changes 
sign  from  positive  through  zero  to  negative,  and  it  follows,  as 
before,  that  its  derivative  <f>IV(x)  is  negative. 

Thus  <f>(a)  is  a  maximum  value  of  <f>(x)  if  <t>'(a)  =  0,  <f>"(a)  =  0, 
<f>'"(a)  =  0,  <j>JV(a)  negative.  Similarly,  <f>(a)  is  a  minimum 
value  of  <f>(x)  if  *'(«)  =  0,  <f>"(a)  =  0,  <£'"(a)  =  0,  and  <£lv(a) 
positive. 

If  it  happens  that  <f>lv(a)  =  0,  it  is  necessary  to  proceed  to 
still  higher  derivatives  to  test  for  turning  values.  The  result 
may  then  be  generalized  as  follows  : 

The  function  <£(#)  has  a  maximum  (or  minimum)  value  at 
x  =  a  if  one  or  more  of  the  derivatives  <£'(a),  <f>"(a),  <f>'"(a)  vanish 
and  if  the  first  one  that  does  not  vanish  is  of  even  order,  and 
negative  (or  positive). 

Ex.  Find  the  critical  values  in  the  example  of  Art.  31  by  the 
second  method. 

<f>"(x)  =  (x+iy(5x-l)+2(x-l)(x+l)(ox-l)  +  5(x-l)(x  +  l)'i 
=  4(5a:8  +  3a:2-3x-l), 

<£"(1)  =  16,  hence  <£(1)  is  a  minimum  value  of  <f>(x)', 
<f>"(—  1)  =  0,  hence  it  is  necessary  to  find  <£'"(—  1) ; 

<£'"(V)  =  12(5  a:2  +  2  a;  -  1), 
<£'"(  —  i.)  —  24,  hence  <£(—  1)  is  neither  a  maximum  nor  a  minimum 
value  of  <j>(x). 

Again,  <£"(£)  =  5(J  -  l)(i  4-  l)2  is  negative,  hence  <£(£)  is  a  maxi- 
mum value  of  <£(V). 


MAXIMA  AND   MINIMA  59 

33.  The  maxima  and  minima  of  any  continuous  function  occur 
alternately.  It  has  been  seen  that  the  maximum  and  minimum 
values  of  a  rational  polynomial  occur  alternately  when  the 
variable  is  continually  increased,  or  diminished. 

This  principle  is  true  also  in  the  case  of  every  continuous 
function  of  a  single  variable.  For,  let  <£(a),  <f>(b)  be  two 
maximum  values  of  <f>(x),  iu  which  a  is  supposed  less  than 
b.  Then,  when  x  =  a  +  h,  the  function  is  decreasing ;  when 
x  =  b  —  h,  the  function  is  increasing,  h  being  taken  sufficiently 
small  and  positive.  But  in  passing  from  a  decreasing  to  an 
increasing  state,  a  continuous  function  must,  at  some  inter- 
mediate value  of  x,  change  from  decreasing  to  increasing,  that 
is,  must  pass  through  a  minimum.  Hence,  between  two  maxima 
there  must  be  at  least  one  minimum. 

It  can  be  similarly  proved  that  between  two  minima  there 
must  be  at  least  one  maximum. 

34.  Simplifications  that  do  not  alter  critical  values.  The  work 
of  finding  the  critical  values  of  the  variable,  in  the  case  of  any 
given  function,  may  often  be  simplified  by  means  of  the  follow- 
ing self-evident  principles. 

1.  When  c  is  independent  of  x,  any  value  of  x  that  gives  a 
turning  value  to  c<jy(x),  gives  a  turning  value  to  <f>(x)  also;  and 
conversely.  These  two  turning  values  are  of  the  same  or 
opposite  kind  according  as  c  is  positive  or  negative. 

2.  Any  value  of  x  that  gives  a  turning  value  to  c  +  <f>(x)  gives 
a  turning  value  of  the  same  kind  to  <f>(x)  also ;  and  conversely. 

3.  When  n  is  independent  of  x,  any  value  of  x  that  gives  a 
turning  value  to  [<£(aj)]B  gives  a  turning  value  to  <f>(x)  also; 
and  conversely.  These  turning  values  are  of  the  same  or 
opposite  kind  according  as  ?i[<£(a,*)]n_1  is  positive  or  negative. 


60  DIFFERENTIAL   CALCULUS 

EXERCISES 

Find  the  critical  values  of  x  in  the  following  functions,  determine 
the  nature  of  the  function  at  each,  and  obtain  the  graph  of  the  function. 

1.  M  =  X(x2  —  1). 

2.  u  =  2  x3  -  15  x2  +  36  x  -  4. 

3.  u  =  (x  -  l)3(x  -  2)2. 

4.  u  =  sin  x  +  cos  x. 

5.  u  =Q-*)3. 

a  -  2x 


7. 

u  =  5  +  12  x  -  3  x1  -  2  x8. 

8. 

logx 

X 

9. 

m  =  sin2  x  cos3  x. 

10. 

^  _  x2  -  X  +  1 

X2  +  X  -  1 

11. 

w_(x  +  3)(x+l) 
(x  -  l)(x-2) 

6.   w  =  x(x  +  l)2  -  5. 

12.  Show  that  a  quadratic  integral  function  always  has  one  maxi- 
mum, or  one  minimum,  but  never  both. 

13.  Show  that  a  cubic  integral  function  has  in  general  both  a 
maximum  and  a  minimum  value,  but  may  have  neither. 

14.  Show  that  the  function  (x  —  b)*  has  neither  a  maximum  nor 
a  minimum  value. 

35.  Geometric  problems  in  maxima  and  minima.  The  theory 
of  the  turning  values  of  a  function  has  important  applications 
in  solving  problems  concerning  geometric  maxima  or  minima, 
i.e.  the  determination  of  the  largest  or  the  smallest  value  a 
magnitude  may  have  while  satisfying  certain  stated  geometric 
conditions. 

The  first  step  is  to  express  the  magnitude  in  question 
algebraically.  Jf  the  resulting  expression  contains  more  than 
one  variable,  the  stated  conditions  will  furnish  enough  relations 
between  these  variables,  so  that  all  the  others  may  be  expressed 
in  terms  of  one.  The  expression  to  be  maximized  or  minimized, 
being  thus  made  a  function  of  a  single  variable,  can  be  treated 
by  the  preceding  rules. 


MAXIMA   AND   MINIMA  61 

Ex.  1.  Find  the  largest  rectangle  whose  perimeter  is  100.  Let  x, 
y  denote  the  dimensions  of  any  of  the  rectangles  whose  perimeter  is 
100.     The  expression  to  be  maximized  is  the  area 

u  =  xy,  (1) 

in  which  the  variables  x,  y  are  subject  to  the  stated  condition 

2x  +  2r/  =  100, 
i.e.  y  =50  -x;  (2) 

hence  the  function  to  be  maximized,  expressed  in  terms  of  the  single 
variable  x,  is  u  _  <£(V)  =  ^50  -x)=  50  x  -  x2.  (3) 

The  critical  value  of  x  is  found  from  the  equation 

<P(x)  =  50-2x  =  0 
to  be  x  =  25.  When  x  increases  through  this  value,  <f>'(x)  changes 
sign  from  positive  to  negative,  and  hence  <f>(x)  is  a  maximum  when 
x  =  25.  Equation  (2)  shows  that  the  corresponding  value  of  y  is  25. 
Hence  the  maximum  rectangle  whose  perimeter  is  100  is  the  square 
whose  side  is  25. 

Ex.  2.  If,  from  a  square  piece  of  tin  whose  side  is  a,  a  square  be 
cut  out  at  each  corner,  find  the  side  of  the  latter  square  in  order  that 
the  remainder  may  form  a  box  of  maximum 
capacity,  with  open  top. 

Let  x  be  a   side   of  each  square  cut  out. 
Then  the  bottom  of  the  box  will  be  a  square 
whose  side  is  a  —  2  x,  and  the  depth  of  the  box 
will  be  x.     Hence  the  volume  is 
v  =  x(a  -  2  x)2, 
which  is  to  be  made  a  maximum  by  varying  x. 

Here  —  =  (a-2i)2^i(a-2z) 

dx 

=  (a-2x)(a  -6.i). 

This  derivative  vanishes  when  x  —  -,  and  when  x  —  -  •     It  will  be 

2  6 

found,  by  applying  the  usual  test,  that  x  =  ^  gives  v  the  minimum 


62 


DIFFERENTIAL   CALCULUS 


value  zero,  and  that  x  =  -  gives  it  the  maximum  value  "  a  •     Hence 

6  27 

the  side  of  the  square  to  be  cut  out  is  one  sixth  the  side  of  the  given 
square. 

Ex.  3.    Find  the  area  of  the  greatest  rectangle  that  can  be  inscribed 

in  a  given  ellipse. 

An  inscribed  rec- 
tangle will  evidently  be 
symmetric  with  regard 
to  the  principal  axes 
of  the  ellipse. 

Let  a,  b  denote  the 
lengths  of  the  semi- 
axes  OA,  OB  (Fig.  14) ; 
let  2  x,  2  y  be  the  dimen- 


Fig.  14 
sions  of  an  inscribed  rectangle. 


Then  the  area  is 

u  =  4  x?j,  (1) 

in  which  the  variables  x,  y  may  be  regarded  as  the  coordinates  of  the 
vertex  P,  and  are  therefore  subject  to  the  equation  of  the  ellipse 


2         bl 


(2) 


It  is  geometrically  evident  that  there  is  some  position  of  P  for 
which  the  inscribed  rectangle  is  a  maximum. 

The  elimination  of  y  from  (1),  by  means  of  (2),  gives  the  function 
of  x  to  be  maximized,  4  j} 


u  =  —  x 
a 


V^ 


(3) 


By  Art.  34,  the  critical  values  of  x  are  not  altered  if  this  function 

is  divided  by  the  constant  — ,  and  then  squared.     Hence,  the  values 

a 

of  x  which  render  u  a  maximum,  give  also  a  maximum  value  to  the 

function  +(x)  =  x*(a*  -  &)  =  aW  -  x*. 

Here  <f>'(x)  =  2  a2x  -  4  xz  =  2  x(a*  -  2  x°-), 

cf>"(x)  =2a2-12r2; 


MAXIMA   AND    MINIMA 


63 


hence,  by  the  usual  tests,  the  critical  values 


±  ■ —  render  cb(x), 

V2 


and  therefore  the  area  w,  a  maximum.     The  corresponding  values  of 
y  are  given  by  (2),  and  the  vertex  P  may  be  at  any  of  the  four  points 


denoted  by 


,     a  ,     b 

±— :,   y  =  ±— , 

V2  V2 


giving  in  each  case  the  same  maximum  inscribed  rectangle,  whose 
dimensions  are  a  V2,  b  V2,  and  whose  area  is  2  ab,  or  half  that  of  the 
circumscribed  rectangle. 

Ex.  4.    Find  the  greatest  cylinder   that  can  be  cut  from  a  given 
right  cone,  whose  height  is  k,  and  the    radius  of   whose  base  is  a. 

Let  the  cone  be  generated  by 
the  revolution  of  the  triangle  0^45 
(Fig.  15),  and  the  inscribed  cylin- 
der be  generated  by  the  revolution 
of  the  rectangle  A  P. 

Let  0 A  =  h,  AB  =a,  and  let  the 

coordinates  of  P  be  (x,  ?j) .     Then 

the  function  to  be  maximized  is 

Try'2(h  r-  x)  subject  to  the  relation  ^  =  - 

x     h 
This  expression  becomes 

v  =  ^.x*a- 
/i2      v 


Fig.  15 


The  critical  value  of  x  is  f  h,  and  V 


x). 

ijraVi 
27 


EXERCISES  ON  CHAPTER  IV 
1.    What  is  the  width  of  the  rectangle  of  maximum  area  that  can 
be  inscribed  in  a  given  right  segment  of  a  parabola? 


.. 


2.    Divide  10  into  two  parts  such  that  the  sum  of  their  squares  is 


a  minimum. 


3.    Find  the  number  that  exceeds  its  square  by  the  greatest  pos- 
sible quantity. 


64  DIFFERENTIAL   CALCULUS 

>^4.    What  number  added  to  its  reciprocal  gives  the  least  possible 
sum? 

5.  Given  the  slant  height  of  a  right  cone;  find  its* altitude  when 
the  volume  is  a  maximum. 

6.  A  rectangular  piece  of  pasteboard  30  in.  long  and  14  in.  wide 
has  a  square  cut  out  at  each  corner.  Find  the  side  of  this  square  so 
that  the  remainder  may  form  a  box  of  maximum  contents. 

7.  Find  the  altitude  of  the  right  cylinder  of  greatest  volume  in- 
scribed in  a  sphere  of  radius  r. 

8.  Determine  the  greatest  rectangle  that  can  be  inscribed  in  a 
given  triangle  whose  base  is  2  b,  and  whose  altitude  is  2  a. 

9.  A  rectangular  court  is  to  be  built  so  as  to  contain  a  given  area 
c2,  and  a  wall  already  constructed  is  available  for  one  of  its  sides. 
Find  its  dimensions  so  that  the  expense  incurred  in  building  the  walls 
for  the  other  sides  may  be  the  least  possible. 

10.  The  volume  of  a  cylinder  of  revolution  being  constant,  find 
the  relation  between  its  altitude  and  the  radius  of  its  base  when  the 
entire  surface  is  a  minimum. 

11.  Assuming  that  the  stiffness  of  a  beam  of  rectangular  cross 
section  varies  directly  as  the  breadth  and  as  the  cube  of  the  depth,  what 
must  be  the  breadth  of  the  stiffest  beam  that  can  be  cut  from  a 
log  16  in.  in  diameter? 

12.  A  man  who  can  row  4  mi.  per  hour,  and  can  walk  5  mi.  per 
hour,  is  in  a  boat  3  mi.  from  the  nearest  point  on  a  straight  beach, 
and  wishes  to  reach  in  the  shortest  time  a  place  on  the  shore  5  mi. 
from  this  point.     Where  must  he  land  ? 

13.  If  the  cost  per  hour  for  the  fuel  required  to  run  a  given 
steamer  is  proportional  to  the  cube  of  her  speed  and  is  $20  an  hour 
for  a  speed  of  10  knots,  and  if  other  expenses  amount  to  $135  an  hour, 
find  the  most  economical  rate  at  which  to  run  her  over  a  course  s. 


MAXIMA   AND   MINIMA  65 

14.  If  the  cost  per  hour  of  running  a  boat  in  still  water  is  propor- 
tional to  the  cube  of  the  velocity,  find  the  most  economical  rate  at  which 
to  run  the  steamer  upstream  against  a  current  of  a  miles  per  hour. 

15.  A  Norman  window  consists  of  a  rectangle  surmounted  by  a 
semicircle.  If  the  perimeter  of  the  window  is  given,  what  must  be  its 
proportions  in  order  to  admit  as  much  light  as  possible? 

16.  Find  the  most  economical  proportions  for  a  cylindrical  dipper 
which  is  to  hold  a  pint. 

17.  The  gate  in  front  of  a  man's  house  is  20  yd.  from  the  car 
track.  If  the  man  walks  at  the  rate  of  4  mi.  an  hour  and  the  car  on 
which  he  is  coming  home  is  running  at  the  rate  of  12  mi.  an  hour, 
where  ought  he  to  get  off  in  order  to  reach  home  as  early  as  possible? 

18.  How  much  water  should  be  poured  into  a  cylindrical  tin  dip- 
per in  order  to  bring  the  center  of  gravity  as  low  down  as  possible  ? 
[Omit  until  after  reading  Art.  164.] 

19.  A  statue  10  ft.  high  stands  on  a  pedestal  that  is  50  ft.  high. 
How  far  ought  a  man  whose  eyes  are  5  ft.  above  the  ground  to  stand 
from  the  pedestal  in  order  that  the  statue  may  subtend  the  greatest 
possible  angle? 

20.  The  sum  of  the  surfaces  of  a  sphere  and  a  cube  is  given.  How 
do  their  dimensions  compare  when  the  sum  of  their  volumes  is  a 
minimum  ? 

21.  An  electric  light  is  to  be  placed  directly  over  the  center  of  a 
circular  plot  of  grass  100  ft.  in  diameter.  Assuming  that  the  inten- 
sity of  light  varies  directly  as  the  sine  of  the  angle  under  which  it 
strikes  an  illuminated  surface  and  inversely  as  the  square  of  its  dis- 
tance from  the  surface,  how  high  should  the  light  be  hung  in  order 
that  the  most  light  possible  shall  fall  on  a  walk  along  the  circumfer- 
ence of  the  plot  ? 

22.  Find  the  relation  between  length  of  circular  arc  and  radius,  in 
order  that  the  area  of  a  circular  sector  of  a  given  perimeter  shall  be  a 
maximum. 

el.  calc.  —  5 


66  DIFFERENTIAL  CALCULUS 

23.  On  the  line  joining  the  centers  of  two  mutually  external 
spheres  of  radii  r,  R,  find  the  distance  of  the  point  from  the  center  of 
the  first  sphere  from  which  the  maximum  of  spherical  surface  is  visible. 

24.  The  radius  of  a  circular  piece  of  paper  is  r.  Find  the  arc  of 
the  sector  which  must  be  cut  from  it  so  that  the  remaining  sector 
may  form  the  convex  surface  of  a  cone  of  maximum  volume. 

25.  Describe  a  circle  with  its  center  on  a  given  circle  so  that  the 
length  of  the  arc  intercepted  within  the  given  circle  shall  be  a  maxi- 
mum. 

26.  Through  a  given  point  within  an  angle  draw  a  straight  line 
which  shall  cut  off  a  minimum  triangle. 

,27.  What  is  the  length  of  the  axis,  and  the  area,  of  the  maximum 
parabola  which  can  be  cut  from  a  given  right  circular  cone,  given 
that  the  area  of  the  parabola  is  equal  to  two  thirds  of  the  product  of 
its  base  and  altitude?  A  parabola  is  cut  from  the  cone  by  a  plane 
parallel  to  an  element. 

28.  Through  the  point  (a,  b)  a  line  is  drawn  such  that  the  part 
intercepted  between  the  rectangular  coordinate  axes  is  a  minimum. 
Find  its  length. 

29.  The  lower  corner  of  a  leaf,  whose  edge  is  a,  is  folded  over  so 
as  just  to  reach  the  inner  edge  of  the  page.  Find  the  width  of  the 
part  folded  over  when  the  length  of  the  crease  is  a  minimum. 

30.  What  is  the  length  of  the  shortest  line  that  can  be  drawn  tan- 
gent to  the  ellipse  b2x2  +  a2y2  =  a2b2  and  having  its  ends  on  the  co- 
ordinate axes  ? 

31.  Given  a  point  on  the  axis  of  the  parabola  y2  =  2  px  at  a  dis- 
tance a  from  the  vertex.  Find  the  abscissa  of  the  point  of  the  curve 
nearest  to  it. 

32.  A  wall  6  ft.  high  is  parallel  to  the  front  of  a  house  and  8  ft. 
from  it.  Find  the  length  of  the  shortest  ladder  that  will  reach  the 
house  if  one  end  rests  on  the  ground  outside  the  wall. 


MAXIMA   AND   MINIMA  67 

33.  It  is  required  to  construct  from  two  circular  iron  plates  of 
radius  a  a  buoy,  composed  of  two  equal  cones  having  a  common  base, 
which  shall  have  the  greatest  possible  volume.  Find  the  radius  of 
the  base. 

34.  A  weight  W  is  to  be  raised  by  means  of  a  lever  with  force  F 
at  one  end  and  the  point  of  support  at  the  other.  If  the  weight  is 
suspended  from  a  point  at  a  distance  a  from  the  point  of  support,  and 
the  weight  of  the  beam  is  w  pounds  per  linear  foot,  what  should  be 
the  length  of  the  lever  in  order  that  the  force  required  to  lift  the 
weight  shall  be  a  minimum? 

35.  A  load  is  hauled  up  an  inclined  plane  by  a  horizontal  force ;  it 
is  required  to  find  the  inclination  0  of  the  plane  so  that  the  mechanical 
efficiency  may  be  greatest,  assuming  that  the  efficiency  77  is  defined  by 
the  formula  tan  Q 


'       tan(0+4>)' 

where  <f>  is  the  angle  of  friction;  i.e.  tan  <f>  —  fi,  the  coefficient  of  fric- 
tion between  the  load  and  the  plane. 

36.  If  the  plane  is  of  cast  iron  and  the  load  is  steel,  and  if  the 
coefficient  of  friction  between  these  substances  is  fi  =  0.347,  at  what 
angle  0  is  the  efficiency  of  the  inclined  plane  a  maximum? 

37.  Prove  that  a  conical  tent  of  given  capacity  will  require  the 
least  amount  of  canvas  when  the  height  is  v2  times  the  radius  of 
the  base. 

38.  If  given  currents  c  and  c'  produce  deflections  a  and  a'  in  a 
tangent  galvanometer,  so  that  tan  «/tan  «'  =  c/c',  show  that  a  —  a'  is 
a  maximum  when  a  +  a'  =  -. 


CHAPTER  V 

RATES  AND  DIFFERENTIALS 

36.  Rates.  Time  as  independent  variable.  Suppose  a  particle 
P  is  moving  in  any  path,  straight  or  curved,  and  let  s  be  the 
number  of  space  units  passed  over  in  t  seconds.  Then  s  may 
be  taken  as  the  dependent  variable,  and  t  as  the  independent 
variable.  The  motion  of  P  is  said  to  be  uniform  when  equal 
spaces  are  passed  over  in  equal  times.  The  number  of  space 
units  passed  over  in  one  second  is  called  the  velocity  of  P, 
The  velocity  v  is  thus  connected  with  the  space  s  and  the  time 

t  by  the  formula  s 

v  —  -. 
t 

The  motion  of  P  is  said  to  be  non-uniform  when  equal  spaces 
are  not  passed  over  in  equal  times.  If  s  is  the  number  of  space 
units  passed  over  in  t  seconds,  then  the  average  velocity  during 
these  t  seconds  is  denned  as  -.  If  during  the  time  A£  the  num- 
ber of  space  units  As  are  described,  then  the  average  velocity 

As 
during  the  time  A£  is  — .     The  actual  velocity  of  P  at  any  in- 
stant  of    time   t    is    the    limit    which    the    average   velocity 
approaches  as  At  is  made  to  approach  zero  as  a  limit. 

rr,  lim    As       ds 

Thus  v  =  A,  .  n —  =  — 

A*  =  °A*      dt 

is  the  actual  velocity  of  P  at  the  time  denoted  by  t.  It  is 
evidently  the  number  of  space  units  that  would  be  passed  over 

68 


RATES  AND   DIFFERENTIALS  69 

in  the  next  second  if  the  velocity  remained  uniform  from  the 
time  t  to  the  time  t  -f  1. 

It  may  be  observed  that  if  the  more  general  term,  "  rate 
of  change,''  is  substituted  for  the  word  "  velocity,"  the  above 
statements  will  apply  to  any  quantity  that  varies  with  the 
time,  whether  it  be  length,  volume,  strength  of  current,  or  any 
other  function  of  the  time.  For  instance,  let  the  quantity  of 
an  electric  current  be  C  at  the  time  t,  and  C  -\-  AC  at  the  time 
t  +  At.     Then  the  average  rate  of  change  of  current  in  the  in- 

terval  At  is  — ;  this  is  the  average  increase  in  current-units 
At 

per  second.     And  the  actual  rate  of  change  at  the  instant  de- 
noted by  t  is  k  rt      jn 
J                             lim    AC      dC 

te±°  At~  dt' 

This  is  the  number  of  current-units  that  would  be  gained  in 
the  next  second  if  the  rate  of  gain  were  uniform  from  the  time 
t  to  the  time  t  +  1.     Since,  by  Art.  8, 

dy_dy.  dx 
dx     dt    dt 

hence  -^  measures  the  ratio  of  the  rates  of  change  of  y  and 

dx 
oi  x. 

It  follows  that  the  result  of  differentiating 

2/=/(*)  (1) 

may  be  written  in  either  of  the  forms 

*-m  (2) 

*-/«!■  (3) 


70  DIFFERENTIAL   CALCULUS 

The  latter  form  is  often  convenient,  and  may  also  be  obtained 
directly  from  (1)  by  differentiating  both  sides  with  regard  to 
t.  It  may  be  read:  the  rate  of  change  of  y  is  f(x)  times  the 
rate  of  change  of  x. 

Ketnrning  to  the  illustration  of  a  moving  point  P,  let  its 

dx 
coordinates  at  time  t  be  x  and  y.     Then        measures   the   rate 

dt 
of  change  of  the  ^-coordinate. 

Since  velocity  has  been  defined  as  the  rate  at  which  a  point 

is  moving,  the  rate  —may  be  called  the  velocity  which  the 
dt 

point  P  has  in  the  direction  of  the  a>axis,  or,  more  briefly,  the 
^-component  of  the  velocity  of  P. 

It  was  shown  on  p.  68  that  the  actual  velocity  at  any  instant 
t  is  equal  to  the  space  that  woiild  be  passed  over  in  a  unit  of 

time,  provided  the  velocity  were 
*lrt\B      /  c^r&-  uniform   during   that   unit.     Ac- 

cordingly,   the    ^-component    of 

dt  velocity  —  may  be  represented 

by   the    distance   PA    (Fig.   16) 
which  P  would  pass  over  in  the 
direction  of  the  a>axis  during  a 
unit  of  time  if  the  velocity  remained  uniform. 

Similarly  -#  is  the  ^/-component  of  the  velocity  of  P,  and 

may  be  represented  by  the  distance  PB. 

The  velocity  —  of  P  along  the  curve  can  be  represented  by 

Civ 

the  distance  PC,  measured  on  the  tangent  line  to  the  curve  at 
P.  It  is  evident  from  the  parallelogram  of  velocities  that  PC 
is  the  diagonal  of  the  rectangle  PA,  PB. 


^.dx 


Fig.  16 


RATES   AND   DIFFERENTIALS  71 

Since  PC2  =  PA2  +  PB2,  it  follows  that 

S)'=(f)"+(*)'  « 

Ex.  1.    If  a  point  describes  the  straight  Hue  3  x  +  4y  =  5,  and  if  x 
increases  h  units  per  second,  find  the  rates  of  increase  of  y  and  of  s. 

Since  y  =  £  -  £  x, 

hence  -7-—  —  7-r" 

rf<  4  eft 

When  —  =  A, 

dt 

it  follows  that  ^  =  -  f  A,  ^  =  VW^TJ^i  -  5  A< 
eft  </£ 


Ex.  Si.  A  point  describes  the  parabola  #2  =  12  r  in  such  a  way  that 
when  x  =  3  the  abscissa  is  increasing  at  the  rate  of  2  ft.  per  second ; 
at  what  rate  is  y  then  increasing?     Find  also  the  rate  of  increase  of  s. 


Since 

y*=12*, 

then 

"(It     '"  dt 

dy     Qdx         6.     dx , 

dt      y  dt      Vl2x  dt ' 

hence    when   x  =  3    and     —  =  2,  it  follows  that  ^  =  ±  2. 

A^-  (D2=  (fT+  B) '  •--  |=^ft.  per  second. 


Ex.  3.  A  person  is  walking  toward  the  foot  of  a  tower  on  a  hori- 
zontal plane  at  the  rate  of  5  mi.  per  hour.  At  what  rate  is  he  ap- 
proaching the  top,  which  is  60  ft.  high,  when  he  is  80  ft.  from  the 
bottom  ? 


72  DIFFERENTIAL   CALCULUS 

Let  x  be  the  distance  from  the  foot  of  the  tower  at  time  t,  and  y 
the  distance  from  the  top  at  the  same  time.     Then 

x2  +  602  =  if, 

and  ***  =  «& 

dt     -dt 

When  x  is  80  ft.,  y  is  100  ft.;    hence  if  —    is   5   mi.   per   hour, 

dy  ■    A      •  ,  dt 

-^  is  4  mi.  per  hour. 

dt  k 

37.  Abbreviated  notation  for  rates.  When,  as  in  the  above 
examples,  a  time  derivative  is  a  factor  of  each  member  of  an 
equation,  it   is   usually  convenient   to    write,  instead  of  the 

symbols  — ,  -^,  the  abbreviations  dx  and  dy,  for  the  rates  of 

J  dt   dt  J' 

change  of  the  variables  x  and  y.     Thus  the  result  of  differen- 

tiatillg  y=m  (l) 

may  be  written  in  either  of  the  forms 

*-/«,  (2) 

dt     '      }  dt'  w 

dy=f'(x)dx.  (4) 

It  is  to  be  observed  that  the  last  form  is  not  to  be  regarded 
as  derived  from  equation  (2)  by  separation  of  the  symbols,  dy, 

dx\  for  the  derivative  -M-  has  been  defined  as  the  result  of 
dx 

performing  upon  y  an  indicated  operation  represented  by  the 

symbol   — ,  and  thus  the  dy  and  dx  of  the  symbol  -M-  have 
dx  dx 

been  given  no  separate  meaning.     The  dy  and  dx  of  equation 

(4)  stand  for  the  rates,  or  time  derivatives,  ~  and   •-—   occur- 

dt         -    dt 


RATES   AND   DIFFERENTIALS  73 

ring  in  (3),  while  the  latter  equation  is  itself  obtained  from 
(1)  by  differentiation  with  regard  to  t,  by  Art.  8. 

In  case  the  dependence  of  y  upon  x  is  not  indicated  by  a 
functional  operation/,  equations  (3),  (4)  take  the  form 


dy 

dy  dx 

dt 

dx  dt ' 

dy  = 

-fax. 

dx 

In  the  abbreviated  notation,  equation  (4)  of  the  last  article 
is  written      (ds)2  —  (dx)2  -f  (dy)2  or  ds2  =  dx2  +  dy2. 

Ex.  1.  A  point  describing  the  parabola  y2  —  2 px  is  moving  at  the 
time  t  with  a  velocity  of  v  ft.  per  second.  Find  the  rate  of  increase 
of  the  coordinates  x  and  y  at  the  same  instant. 

Differentiating  the  given  equation  with  regard  to  t,  we  obtain 
ydy  =  pdx. 

But  dx,  dy  also  satisfy  the  relation 
dx2  +  dy2  =  v2 ; 

hence,  by  solving  these  simultaneous  equations,  we  obtain 

dx  =  —    '  v,         dy  =        ^         v,  in  feet  per  second. 

v V  +  p2  Vy*2  +  p2 

Ex.  2.  A  vertical  wheel  of  radius  10  ft.  is  making  5  revolutions  per 
second  about  a  fixed  axis.  Find  the  horizontal  and  vertical  velocities 
of  a  point  on  the  circumference  situated  30°  from  the  horizontal. 

Since  x  =  10  cos  6,         y  =  10  sin  6, 

then  dx  =  -10  sin  OdO,         dy  =  10  cos  Odd. 

But  dO  =  10  7T  =  31.416  radians  per  second, 

hence  dx  =  —  314.16  sin  $  =  —  157.08  ft.  per  second, 

and  dy  =  314.16  cos  0  =  272.06  ft.  per  second. 

Ex.  3.  Trace  the  changes  in  the  horizontal  and  vertical  velocity 
in  a  complete  revolution. 


74 


DIFFERENTIAL   CALCULUS 


38.  Differentials  often  substituted  for  rates.  The  symbols  dx, 
dy  have  been  denned  above  as  the  rates  of  change  of  x  and  y 
per  second. 

Sometimes,  however,  they  may  conveniently  be  allowed  to 
stand  for  any  two  numbers,  large  or  small,  that  are  propor- 
tional to-  these  rates ;  the  equations,  being  homogeneous  in 
them,  will  not  be  affected.  It  is  usual  in  such  cases  to  speak 
of  the  numbers  dx  and  dy  by  the  more  general  name  of  differ- 
entials ;  they  may  then  be  either  the  rates  themselves,  or  any 
two  numbers  in  the  same  ratio. 

This  will  be  especially  convenient  in  problems  in  which  the 
time  variable  is  not  explicitly  mentioned. 

39.  Theorem  of  mean  value.  Let  f(x)  be  a  continuous  func- 
tion of  x  which  has  a  derivative.     It  can  then  be  represented 

by  the  ordinates  of  a  curve  whose 
equation  is  y  —f(x). 
In  Pig.  17,  let 

x  =  ON,   x  +  h  =  OR, 
f(x)  =  NH,  f(x  +  h)  =  RK. 
Then  f(x  +  li)-f(x)  =  MK,  and 
f(x+iC\-f(xS     MK 


Fig.  17 


=  ^±±  =  tan  MHK. 


h  HM 

But  at  some  point  8  between  H  and  A"  the  tangent  to  the 
curve  is  parallel  to  the  secant  IIK.  Since  the  abscissa  of  S  is 
greater  than  x  and  less  than  x  +  h  it  may  be  represented  by 
x-\-0h,  in  which  6  is  a  positive  number  less  than  unity.  The 
slope  of  the  tangent  at  S  is  then  expressed  by/'(.x*  +  6h),  hence 
f(x  +  h)  -f(x)  =  /{x  +  m)t 


from  which 


f(x  +  h)=f(x)+hf'(x+Oh). 


RATES   AND   DIFFERENTIALS  75 

The  theorem  expressed  by  this  formula  is  known  as  the 
theorem  of  mean  value. 

If  in  this  equation  we  put 

f(x  +  K)  —  f  (a?)  =  dy,         h  =  dx, 

in  which  h  is  an  arbitrary  increment,  then  the  relation  between 
the  increment  of  the  variable  and  the  actual  increment  of  the 
function  will  be  expressed  by  the  equation 

dy=f'(x  +  6dx)dx, 

whereas  if  dy,  dx  are  regarded  as  differentials  (dy  not  an 
actual  but  a  virtual  increment),  then  the  relation  becomes 

dy=f\x)dx. 

This  more  clearly  illustrates  that  the  differential  dy  is  de- 
fined as  the  change  that  would  take  place  in  the  function  y, 
corresponding  to  the  actual  change  dx  in  the  independent  vari- 
able x}  provided  the  rate  of  change  remained  constant. 

EXERCISES 

1.  When  x  increases  from  45°  to  45°  15',  find  the  increase  of 
logio  sin  x,  assuming  that  the  ratio  of  the  rates  of  change  of  the  func- 
tion and  the  variable  remains  constant  throughout  the  short  interval. 

Here  dy  =  logio  e  •  cot  xdx  =  .4343  cot  xdx  =  .4343  dx. 

Let  dx—  .004163  (the  number  of  radians  in  15'). 

Then  dy  =  .001895, 

which  is  the  approximate  increment  of  logio  sin  x. 
But  log10  sin  45°  =  -  \  log  2  =  -  .150515, 

therefore  log10  sin  45°  15'  =  -  .148620. 

2.  Show  that  log,ft.r  increases  more  slowly  than  x,  when 
x  >  log10  e,  that  is,  x  >  0.4343. 


76  DIFFERENTIAL   CALCULUS 

3.  A  man  is  walking  at  the  rate  of  5  mi.  per  hour  towards  the 
foot  of  a  tower  60  ft.  high  standing  on  a  horizontal  plane.  At  what 
rate  is  the  angle  of  elevation  of  the  top  changing  when  he  is  80  ft.  from 
the  foot  of  the  tower  ? 

4.  An  arc  light  is  hung  12  ft.  directly  above  a  straight  horizontal 
walk  on  which  a  man  5  ft.  in  height  is  walking.  How  fast  is  the  man's 
shadow  lengthening  when  he  is  walking  awTay  from  the  light  at  the 
rate  of  168  ft.  per  minute? 

5.  At  what  point  on  the  ellipse  16  x2  +  9  y2  —  400  does  y  decrease 
at  the  same  rate  that  x  increases? 

6.  A  vessel  is  sailing  northwest  at  the  rate  of  10  mi.  per  hour. 
At  what  rate  is  she  making  north  latitude? 

7.  In  the  parabola  y2  =  12  x,  find  the  point  at  which  the  ordinate 
and  abscissa  are  increasing  equally. 

8.  At  what  part  of  the  first  quadrant  does  the  angle  increase  twice 
as  fast  as  its  sine  ? 

9.  Find  the  rate  of  change  in  the  area  of  a  square  when  the  side 
b  is  increasing  at  a  ft.  per  second. 

10.  In  the  function  y  =  2  xz  +  6,  what  is  the  value  of  x  at  the  point 
where  y  increases  24  times  as  fast  as  x  ? 

11.  A  circular  plate  of  metal  expands  by  heat  so  that  its  diam- 
eter increases  uniformly  at  the  rate  of  2  in.  per  second.  At  what  rate 
is  the  surface  increasing  when  the  diameter  is  5  in.? 

12.  What  is  the  value  of  x  at  the  point  at  which  x8  —  5  x'2  +  17  x  and 
xs  —  3  x  change  at  the  same  rate  ? 

13.  Find  the  points  at  which  the  rate  of  change  of  the  ordinate 
y  =  x3  -  6  x2  +  3  x  +  5  is  equal  to  the  rate  of  change  of  the  slope  of  the 
tangent  to  the  curve. 

14.  The  relation  between  s,  the  space  through  which  a  body  falls, 


rates  and  differentials  77 

and  t,  the  time  of  falling,  is  s  =  16  t2.     Show  that  the  velocity  is  equal 
to  32  t. 

The  rate  of  change  of  velocity  is  called  acceleration  and  is  denoted 

by  a. 

TT  dv      d2s 

Hence  a  =  —  =  -— • 

dt      dt2 

Show  that  the  acceleration  of  the  falling  body  is  a  constant. 

15.  A  body  moves  according  to  the  law  s  —  cos  (nt  +  e).  Show  that 
its  acceleration  is  proportional  to  the  space  through  which  it  has 
moved. 

16.  If  a  body  is  projected  upwards  in  a  vacuum  with  an  initial 
velocity  v0,  to  what  height  will  it  rise,  and  what  will  be  the  time 
of  ascent? 

17.  A  body  is  projected  upwards  with  a  velocity  of  a  ft.  per  second. 
After  what  time  will  it  return? 

18.  If  A  is  the  area  of  a  circle  of  radius  x,  show  that  the  circum- 

dA 
ference  is  — .     Interpret  this  fact  geometrically. 

19.  A  point  describing  the  circle  x2  +  y2  =  25  passes  through  (3,4) 
with  a  velocity  of  20  ft.  per  second.     Find  its  component  velocities 

parallel  to  the  axes. 

20.  Let  a  point  P  move  with  uniform  velocity  on  a  circle  of  radius 
a  with  center  0 ;  let  AB  be  any  diameter,  and  Q  the  orthogonal  projec- 
tion of  P  on  AB.  Find  an  expression  for  the  velocity  of  Q  in  terms 
of  the  angular  velocity  of  P,  and  show  how  this  velocity  varies  during 
a  revolution  of  P.  The  motion  of  the  point  Q  along  AB  is  called 
harmonic. 

21.  A  point  P  moves  along  the  curve  y  =  xz  at  the  rate  of  3  ft.  per 
second.  At  what  rate  is  the  angle  <£,  which  the  tangent  to  the  curve 
makes  with  the  x-axis,  increasing  when  P  is  passing  through  the 
point  (1,  1)? 


V 


CHAPTER   VI 


DIFFERENTIAL  OF  AN  AREA,   ARC,  VOLUME,   AND 
SURFACE  OF  REVOLUTION 

40.  Differential  of  an  area.  If  the  coordinates  of  P  are  (x,  y) 
and  those  of  Q  (x  4-  Ax,  y-\-  Ay),  then 
MN=  PR  =  Ax,  and  PS  =  RQ=Ay. 
If  the  area  OAPM  is  denoted  by  A, 
then  A  is  evidently  some  function 
of  the  abscissae;  also  if  area  OAQN 
is  denoted  by  A  4-  AA  then  the 
area  MNQP  is  AA ;  it  is  the  incre- 
ment taken  by  the  function  A,  when 

x  takes    the    increment   Ax.      But  MNQP  lies  between  the 

rectangles  MR,  MQ ;  hence 


i/Aa;  <  A.4  <  (y  +  Ay)Aa;, 


and 


A. 4 

y<—-<y  +  ^y- 

Ax 


Therefore,  when  Ax,  Ay,  AA  all  approach  zero, 


lim 


A  A      dA 


Ax 


(lx 


=  ?/ 


Hence,  if  the  ordinate  and  the  area  are  expressed  each  as  a 
function  of  the  abscissa,  the  derivative  of  the  area  function 
with  regard  to  the  abscissa  is  equal  to  the  ordinate  function. 

78 


AREA,  ARC,  VOLUME,  AND  SURFACE 


79 


In  the  notation  of  differentials  we  may  say :  The  differential 
of  the  area  between  a  curve  and  the  axis  of  x  is  measured  by  the 
product  of  the  ordinate  and  the  differential  of  x. 

dA  =  ydx. 

Ex.     If  the  area  included  between  a  curve,  the  axis  of  x,  and  the 

ordinate  whose  abscissa  is  a:,  is  given  by  the  equation 

A  =  xs, 
find  the  equation  of  the  curve. 

Here 


=M  =  3z2. 


dx 

41.  Differential  of  an  arc.  A  segment  of  a  straight  line  is 
measured  by  applying  the  unit  of  measure  successively  to  the 
segment  to  be  measured.  In  the  case  of  a  curve  this  is  gen- 
erally impossible.  We  define  the  length  of  a  given  curve 
between  two  points  upon  it  as  the  limit  of  the  sum  of  the 
chords  joining  points  on  the  curve  when  the  lengths  of  these 
chords  approach  the  limit  zero.  We  shall  then  assume  that  the 
ratio  of  the  arc  to  the  chord  approaches  the  limit  1  when  the 
length  of  the  chord  approaches  the  limit  zero.     [Compare  §  19.] 

Let  PQ  be  two  points  on  the  curve  (Fig.  19) ;  let  x,  y  be  the 
coordinates  of  P ;  x  +  Ax,  y  +  Ay  Y 
those  of  Q ;  s  the  length  of  the  arc 
AP;  s  -+-  As  that  of  the  arc  AQ. 
Draw  the  ordinates  MP,  NQ ;  and 
draw  PR  parallel  to  MN.  Then 
PR  =  Ax,  RQ  =  Ay ;  arc  PQ  =  As.    - 


Hence  chord  PQ  =  V(A  xf+  (Ay)2, 


M     N 
Fig.  19 


PQ 

Ax 


Therefore       '  = 


Ai 

Ax 


As 
1>Q 


AXy 

PQ=  As 
Ax      PQ 


A?/Y 


vi+is- 


80  DIFFERENTIAL   CALCULUS 


Taking  the  limit  of   both  members  as  Ax  approaches  zero 

lim      As 


and  putting      ™  Q  -~  =  1,  we  obtain 


%-Mif 

(1) 

Si,u„„ly,                 |-V'+(*/. 

(2) 

Moreover,  from  Art.  36, 

(ds\*/dx\*    (dy\* 
[dtj      \dtj      \dtj' 

(3) 

or  in  the  differential  notation, 

ds2  =  dx2  -f  dy2. 

W 

42.    Trigonometric  meaning  of  — ,  — . 

doc    dy 

cs-                   A.r      A.r      PQ             »r>/^ 
Since             —  = •  — -3S  =  cos  PPQ 

As      PQ       As 

As' 

it  follows  by  taking  the  limit  that 

dx  , 

—  =  cos  <£, 
ds 

wherein  <f>,  being  the  limit  of  the  angle  RPQ,  is  the  angle 
which  the  tangent  at  the  point  (x,  y)  makes  with  the  o-axis. 

Similarly,  -*  =  sin  </> ;  whence  —  =  sec  <£,  —  =  esc  <£. 
ds  dx  dy 

By  using  the  idea  of  a  rate  or 
differential,  all  these  relations  may 
be  conveniently  exhibited  by  Fig. 
20. 

These  results  may  also  be  de- 
rived   from   equations    (1),  (2)  of 

Art.  41,  by  putting     -'^tan  d>. 
Fig.  20  J  F  dx 


*■     ■ .» 


QL^dx 


AREA,  ARC,  VOLUME,  AND  SURFACE 


81 


43.  Differential  of  the  volume  of  a  solid  of  revolution.  Let 
the  curve  APQ  (Fig.  21)  revolve  about  the  x-axis,  and  thus 
generate  a  surface  of  revo- 
lution ;  let  V  be  the  volume  in- 
cluded between  this  surface, 
the  plane  generated  by  the 
fixed  ordinate  at  A,  and  the 
plane  generated  by  any  ordinate 
MP. 

Let  A  V  be  the  volume  gener- 
ated by  the  area  PMNQ.  Then  AV  lies  between  the  vol- 
umes of  the  cylinders  generated  by  the  rectangles  PMNR 
and  SMNQ)  that  is, 

Try9  Ax  <  A  V  <  7T  (y  -f  Ay)2  Ax. 


Y 

s 

A 

/ 

R 

X 

0 

M    A 

J 

Fig.  21 


Dividing  by  Ax  and  taking  limits,  we  obtain 
dV 


dx 


Try2,     dV  =  7ry2dx. 


44.  Differential  of  a  surface  of  revolution.  Let  S  be  the  area 
of  the  surface  generated  by  the  arc  AP  (Fig.  22),  and  AS  that 
generated  by  the  arc  PQ,  whose  length  is  As. 

Draw  PQ',  QP  parallel  to  OX 
and  equal  in  length  to  the  arc  PQ. 
Then  it  may  be  assumed  as  an 
axiom  that  the  area  generated  by 
PQ  lies  between  the  areas  gen- 
iL    erated  by  PQ'  and  P'Q;   i.e. 

2  iry As  <  AS  <  2  tv  (y  +  Ay)  As. 


I 


M      N 
Fig.  22 

EL.    CALC.  —  0 


DIFFERENTIAL   CALCULUS 


Dividing  by  As  and  passing  to  the  limit, 


f  =  2^  (1) 


dx      ds     dx  \        \dx) 


45.    Differential  of  arc  in  polar  coordinates.     Let  p,  6  be  the 
coordinates  of  P  (Fig.  23) ;  p  ■+  Ap,  6  +  A6  those  of  $ ;  s  the 

length  of  the  arc  KP ;  As  that  of  the 
L   arcPQ;  draw  PM  perpendicular  to 
OQ.     Then 
PM  =  p  sin  A0, 


p 

MQ  =  OQ-OM=p  +  AP-p  cos  A<9 

=  p  (1  —  cos  A0)  +  A/a 
Fig.  23  =  2  P  sin2  i  A#  +  V 

Hence      PQ2  =  (p  sin  A0)2  +  (2  p  sin2 1  A0  +  Ap)2, 

/PQV  =   ^sinA^V  +  (p  sin  J  A0  .  SAi^  +  V)\ 

Replacing  the  first  member  by  f  — - %■  •  -^  ] ,  passing  to  the 

\    .AN  -Af7  / 

limit  when   A0  =  0,   and    putting   lira— ^  =  1,   lira  —      —  =  1, 

As  A9 

lira  sin  *  A6  =  1,  we  obtain 
i^0 

^V  =  ,2  +  ^Y. 


that  is,  ^  =  V^+/ 


d<9  '  Vd0y 


AREA,  ARC,  VOLUME,  AND  SURFACE 


83 


In  the  rate  or  differential  notation  this  formula  may  be 
conveniently  written     ^  =  rf  *  +   2^ 

46.  Differential  of  area  in  polar  coordinates.  Let  A  be  the 
area  of  OKP  (Fig.  24)  measured  from  a  fixed  radius  vector 
OK  to  any  other  radius  vec- 
tor OP ;  let  A^4  be  the  area  of 
OPQ.  Draw  arcs  PM,  QN, 
with  0  as  a  center.  Then  the 
area  POQ  lies  between  the 
areas  of  the  sectors  OPM  and 
ONQ;  i.e.  Fig.  24 

1  p2A0  <  A.-l  <  J  (p  +  Ap)2A0. 

Dividing  by  A0  and  passing  to  the  limit,  when  A0  =  0,  we 
obtain 


dA 

d$ 


=  ip2- 


Hence,  in  the  differential  notation  we  may  write  the  formula 
dA  =  \  phlO. 


EXERCISES  ON   CHAPTER  VI 

1.  In  the  parabola  y2  =  4  ax,  find  —  ,   —  ,  — - , 

dx      dx      dx      dx 

2.  Find  —  and  —  for  the  circle  x2  +  y2  =  a2. 

dx  dy 

ds 

3.  Find  —  for  the  curve  ev  cos  x  =  1. 

dx 

4.  Find  the  x-derivative  of  the  volume  of  the  cone  generated  by 
revolving  the  line  y  =  ax  about  the  axis  of  x. 

5.  Find  the  it-derivative  of  the  volume  of  the  ellipsoid  of  revolu- 

tion,  formed  bv  revolving  —  +  •*-  =  1  about  its  major  axis. 
a2      b2 


84  DIFFERENTIAL   CALCULUS 

ds 

6.  In  the  curve  p  =  a9  find  — 

du 

7.  Given  p  =  a  (1  +  cos  6)  ;  find  ^ 

8.  In  p2  cos  2  0,  find  ^. 

9.  The  parabolic  arc  y2  =  9  x  measured  from  the  vertex  to  a  variable 
point  P  =  (x,  y)  is  revolving  about  the  a>axis.  If  P  moves  along  the 
curve  at  the  rate  of  2  in.  per  second,  what  is  the  rate  of  increase 
of  the  surface  of  revolution  when  P  is  passing  through  the  point 
(4,  6)?     What  is  the  rate  of  increase  of  the  volume  of  revolution? 

10.  The  radius  vector  to  the  cardioid  p  =  2  (1  —  cos  $)  is  rotating 
about  the  origin  with  an  angular  velocity  of  18°  per  second.  Find 
the  rate  at  which  the  extremity  P  of  the  radius  vector  is  moving  along 
the  curve,  taking  the  inch  as  unit  of  length.  At  what  points  of  the 
curve  will  P  be  moving  fastest?  slowest?  Find  the  velocities  at 
these  points. 


CHAPTER   VII 

APPLICATIONS  TO  CURVE  TRACING 

47.  Equation  of  tangent  and  normal.  The  function  y=f(x) 
may  be  represented  by  a  plane  ciu-ve.  It  will  now  be  shown 
how  to  obtain  several  of  the  properties  of  this  curve  by  means 
of  the  principles  already  established.  The  tangent  line  at  a 
point  (x-t,  y{)  on  the  curve  passes  through  the  point  and  has 

the  slope  — ,  the  symbol  meaning  that  the  coordinates  xx,  yx 

are  substituted  in  the  first  derivative  after  the  differentiation 
has  been  performed.     Its  equation  may  be  written  in  the  form 

y- 2/1  =  ^' (*-*.)•  (1) 

The  normal  to  the  curve  at  the  point  (xu  yY)  is  the  straight 
line  through  this  point,  perpendicular  to  the  tangent.  Since 
the  slope  of  the  normal  is  the  negative  reciprocal  of  that  of 
the  tangent,  its  equation  may  be  written  in  the  form 

*-*i  +  p(*-  *)-0.  (2) 

48.  Length  of  tangent,  normal,  subtangent,  subnormal.  The 
segments  of  the  tangent  and  normal  intercepted  between  the 
point  of  tangency  and  the  axis  OX  are  called,  respectively, 
the  tangent  length  and  the  normal  length,  and  their  projections 
on  OX  are  called  the  subtangent  and  the  subnormal. 

85 


DIFFERENTIAL  CALCULUS 
C 


Fig.  25  a 


Fig.  25  6 


Thus,  in  Fig.  25,  a,  b  let  the  tangent  and  normal  to  the  curve  PC 
at  P  meet  the  axis  OX  in  T  and  N,  and  let  i/P  be  the  ordi- 
nate of  P.     Then    TP  is  the  tangent  length, 
PNthe  normal  length, 
TM  the  subtangent, 
MN the  subnormal. 
These  will  be  denoted,  respectively,  by  t,  n,  r,  v. 

Let  the  angle  XTP  be  denoted  by  cf>,  and  write  tan  <f>  =  -^*. " 


Then 


hence 


d^x 


2/i 


1  + 


(for, 


y 


H§, 


fdjh 

\dxx 


The  subtangent  is  measured  from  the  intersection  of  the 
tangent  to  the  foot  of  the  ordinate ;  it  is  therefore  positive 
when  the  foot  of  the  ordinate  is  to  the  right  of  the  intersec- 
tion of  tangent.  The  subnormal  is  measured  from  the  foot 
of  the  ordinate  to  the  intersection  of  normal,  and  is  positive 
when  the  normal  cuts  OX  to  the  right  of  the  foot  of  the  ordi- 


APPLICATIONS  TO   CURVE   TRACING  87 

nate.     Both  are  therefore  positive  or  negative,  according   as 
<£  is  acute  or  obtuse. 

The  expressions  for  t,  v  may  be  obtained  also  by  finding 
from  equations  (1),  (2),  Art.  47,  the  intercepts  made  by  the 
tangent  and  normal  on  the  axis  OX.  The  intercept  of  the 
tangent  subtracted  from  xl  gives  t,  and  xx  subtracted  from 
the  intercept  of  the  normal  gives  v. 

Ex.  Find  the  intercepts  made  upon  the  axes  by  the  tangent  at  the 
point  (xv  yx)  on  the  curve  Va:  +  Vy  =  Va,  and  show  that  their  sum 
is  constant. 

Differentiating  the  equation  of  the  curve,  we  obtain 

2  V*     2V~ydx 
Hence  the  equation  of  the  tangent  is 

xi 

The  x  intercept  is  xl  +  y/xlyv  and  the  y  intercept  is  yx  4-  y/xlyv 
hence  their  sum  is  ,   ,—        ,— N0 

If  a  series  of  lines  is  drawn  such  that  the  sum  of  the  intercepts  of 
each  is  the  same  constant,  account  being  taken  of  the  signs,  the  form 
of  the  parabola  to  which  they  are  all  tangent  can  be  readily  seen. 

EXERCISES 

1.  Find  the  equations  of  the  tangent  and  the  normal  to  the  ellipse 

X2         V2 

(-  V-  —  1  at  the  point  (xv  ?/,).     Compare  the  process  with  that  era- 

a'2       b2      i 

ployed  irr-analytic  geometry  to  obtain  the  same  results. 

2.  Find  the  equation  of  the  tangent  to  the  curve 

x\x  +y)=  a2(x -y) 
at  the  origin. 


88  DIFFERENTIAL   CALCULUS 

3.  Find  the  equations  of  the  tangent  and  normal  at  tho  point 
(1,  3)  on  the  curve  y'2  =  9  x3. 

4.  Find  the  equations  of  the  tangent  and  normal  to  each  of  the 
following  curves  at  the  point  indicated  : 

(a)   y  = — ,  at  the  point  for  which  x  =  2  a. 

4  a2  +  x2 

(jg)  y2  =  2  x2  —  x3,  at  the  points  for  which  x  =  1. 

(y)   y2  ~  ipx,  at  the  point  (p,  2p). 

5.  Find  the  value  of  the  subtangent  of  y2  =  3  x2  —  12  at  x  =  4. 
Compare  the  process  with  that  given  in  analytic  geometry. 

6.  Find  the  length  of  the  tangent  to  the  curve  y2  =  2  x  at  x  =  8. 

7.  Find  the  points  at  which  the  tangent  is  parallel  to  the  axis 
of  x,  and  at  which  it  is  perpendicular  to  that  axis  for  each  of  the  fol- 
lowing  curves  :  (ft)   a,?  +  2  hxy  +  bf  =  1. 

(/J)  ,  =?^iL\ 
ax 

(■y)   y3  =  x2(2  a  —  x). 

8.  Find  the  condition  that  the  conies 

ax2  4-  by2  =  1,  a'x2  -f  b'y2  —  1 
shall  cut  at  right  angles. 

9.  Find  the  angle  at  which  x2  =  y2  +  5  intersects  8  x2  -f  18  y2  =  144. 
Compare  with  Ex.  8. 

10.  Show  that  in  the  equilateral  hyperbola  2  xy  —  a2  the  area  of 
the  triangle  formed  by  a  variable  tangent  and  the  coordinate  axes  is 
constant  and  equal  to  a2. 

11.  At  what  angle  does  y2  =  8  x  intersect  4  x2  +  2  y2  =48  ? 

12.  Determine  the  subnormal  to  the  curve  yn  —  aP-^x. 

13.  Find  the  values  of  x  for  which  the  tangent  to  the  curve 

7/3  =(x-  a)2(x  -  c) 
is  parallel  to  the  axis  of  x. 


APPLICATIONS  TO   CURVE   TRACING  89 

14.  Show  that  the  subtangent  of  the  hyperbola  xy  =  a2  is  equal  to 
the  abscissa  of  the  point  of  tangency,  but  opposite  in  sign. 

15.  Prove  that  the  parabola  y2  =  4  ax  has  a  constant  subnormal. 

16.  Show  analytically  that  in  the  curve  x2  +  y2  =  a2  the  length  of 
the  normal  is  constant. 

17.  Show  that  in  the  tractrix,  the  length  of  the  tangent  is  con- 
stant, the  equation  of  the  tractrix  being 


x  =  Vc*-Tj*  +  £  log  c-Vc2-y\ 

-         c  +  vV2  —  y2 

X 

18.  Show  that  the  exponential  curve  y  =  aec  has  a  constant  sub- 
tangent. 

19.  Find  the  point  on  the  parabola  y2  =  4 px  at  which  the  angle 
between  the  tangent  and  the  line  joining  the  point  to  the  vertex  shall 
be  a  maximum. 

49.  Concavity  upward  and  downward.  A  curve  is  said  to  be 
concave  downward  in  the  vicinity  of  a  point  P  when,  for  a 
finite  distance  on  each  side  of  P,  the  curve  is  situated  below 


the  tangent  drawn  at  that  point,  as  in  the  arcs  AD,  FTL  It 
is  concave  upward  when  the  curve  lies  above  the  tangent,  as 
in  the  arcs  DF,  HK. 


90  DIFFERENTIAL   CALCULUS 

By  drawing  successive  tangents  to  the  curve,  as  in  the  fig- 
ure, we  easily  see  that  if  the  point  of  contact  advances  to  the 
right,  the  tangent  swings  in  the  positive  direction  of  rotation 
when  the  concavity  is  upward,  and  in  the  negative  direction 
when  the  concavity  is  downward.  Hence  upward  concavity 
may  be  called  a  positive  bending  of  the  curve,  and  downward 
concavity,  a  negative  bending. 

A  point  at  which  the  direction  of  bending  changes  con- 
tinuously from  positive  to  negative,  or  vice  versa,  as  at  F  or 
at  D,  is  called  a  point  of  inflexion,  and  the  tangent  at  such  a 
point  is  called  a  stationary  tangent. 

The  points  of  the  curve  that  are  situated  just  before  and  just 
after  the  point  of  inflexion  are  thus  on  opposite  sides  of  the 
stationary  tangent,  and  hence  the  tangent  crosses  the  curve,  as 
at  D,  F,  H. 

50.  Algebraic  test  for  positive  and  negative  bending.  Let  the 
inclination  of  the  tangent  line,  measured  from  the  positive  end 
of  the  ^axis  toward  the  forward  end  of  the  tangent,  be  denoted 
by  <£.  Then  <£  is  an  increasing  or  decreasing  function  of  the 
abscissa  according  as  the  bending  is  positive  or  negative ;  for 

instance,   in  the   arc  AD,  the  angle   <f>  diminishes  from  4-  — 

Z 

through   zero  to   —  - ;    in  the  arc  DF,  cfy  increases  from  —  - 
4  4 

through  zero  to  - ;  in  the  arc  FIT,  <£  decreases  from  -f  -  through 
o  o 

zero  to  —  y  ;  and  in  the  arc  HK,  <f>  increases  from  —  y  through 

zero  to  +  -. 

At  a  point  of  inflexion  $  has  evidently  a  turning  value  which 
is  a  maximum  or  a  minimum,  according  as  the  concavity  changes 
from  upward  to  downward,  or  conversely. 


APPLICATIONS  TO  CURVE  TRACING  91 

Thus  in  Fig.  26,  cf>  is  a  maximum  at  F,  and  a  minimum  at  D 
and  at  II. 

Instead  of  recording  the  variation  of  the  angle  <f>,  it  is  gen- 
erally convenient  to  consider  the  variation  of  the  slope,  tan  <£, 
which  is  easily  expressed  as  a  function  of  x  by  the  equation 

tan  <£  =  -^. 
dx 

Since  tan  <f>  is  always  an  increasing  function  of  <f>,  it  follows 

that  the  slope  function  -^  is  an   increasing  or  a  decreasing 

dx 

function  of  x,  according  as  the  concavity  is  upward  or  down- 
ward, and  hence  that  its  x-derivative  is  positive  or  negative. 
Thus  the  bending  of  the  curve  is  in  the  positive  or  negative 

d-y 
direction  of  rotation,  according  as  the  function  — ^  is  positive 

dx- 
or  negative. 

At   a   point   of   inflexion   the   slope  —  is  a  maximum  or  a 

dx 

d2v 
minimum,  and  therefore  its  derivative  — -  changes  sign  from 

dx2 

positive  to  negative  or  from  negative  to  positive.     This  latter 

condition  is  evidently  both  necessary  and  sufficient  in  order  that 

the  point  (x,  y)  may  be  a  point  of  inflexion  on  the  given  curve. 

Hence,  the  coordinates  of  the  points  of  inflexion  on  the  curve 

may  be  found  by  solving  the  equations 

/"O)  =  o,  /"(*)=«,, 

and  then  testing  whether  f"(x)  changes  its  sign  as  x  passes 
through  the  critical  values  thus  obtained.  To  any  critical 
value  a  that  satisfies  the  test  corresponds  the  point  of  inflexion 


92 


DIFFERENTIAL   CALCULUS 


Ex.  1.   For  the  curve        y  =  (x2  —  l)2 
find  the  points  of  inflexion,  and  show  the  mode  of  variation  of  the 
slope  and  of  the  ordinate. 

Here 


^/  =  4a-(>2-l), 


dx 
(Py_ 


dx2 


=  4(3:r2-l), 


1 


hence  the  critical  values  for  inflexions  are  x  =  ± — -.     It  will  be  seen 

i  vs 

that  as  x  increases  through =,  the  second  derivative  changes  sign 

V3 

from   positive    to   negative,    hence    there   is   an    inflexion    at   which 
the  concavity  changes   from   upward  to   downward.      Similarly,  at 

x  =  ^ — =3  the  concavitv  changes  from  downward  to  upward.      The 

V3 
following  numerical  table  will  help  to  show  the  mode  of  variation  of 
the  ordinate  and  of  the  slope,  and  the  direction  of  bending. 

As    x    increases    from     —  oo  to 

=  the  bending  is  positive,  and 

Va 

the  slope  continually  increases  from 
—  go  through  zero  to  a  maximum 

I      Q 

value  — — ,   which  is  the   slope  of 

3V3 
the   stationary  tangent   drawn   at 

the  point  f ,  -  ). 

V      V3   9/ 

As  x  continues  to  increase  from 

L_  to  H =,  the  bending  is  neg- 

V3  V3 

Q 

ative,  and  the  slope  decreases  from  +  — -  through  zero  to  a  minimum 

oV3 

o 

value _,  which  is  the  slope  of  the  stationary  tangent  at 

3V<3 


X 

y 

dy 

dx 

d*y 
dx2 

—  00 

+  co 

—   GO 

+ 

_  2 

+  9 

-24 

+ 

-1 

0 

0 

+ 

V3 

-1 

8 
3V3 

0 

0 

1 

0 

_ 

V3 

H 

8 
3V3 

0 

1 

0 

0 

+ 

+  CO 

+  00 

+  GO 

+ 

^■tl 


APPLICATIONS  TO   CURVE   TRACING 


93 


Finally,  as  x  increases  from  -\ to  +  oo,  the  bending  is  positive 

V3 
and  the  slope  increases   from  the 

o 

value through  zero  to  +  oo. 

3V3 
The   values   x  =  —  1,  0,  +1,   at 
which  the  slope  passes  through  zero, 
correspond  to  turning  values  of  the 
ordinate. 


Ex.  2.    Examine   for   inflexions 

the  curve 

x  +  4  =  (y  -  2)3. 


V 


J 


Fig.  27 


In  this  case 

y  =  2+(x  +  4)\ 

dy      1 
dx 
d2y 


(x  +  4)" 


dx2 


0  +  4) 


dy 


Fig.  28 


Hence,  at  the  point  (  -  4,  2),  ^ 

dx 

and  — ¥  are  infinite.   When  x<  -  4, 
dx'2 


d2u  d2v 

—±  is  positive,  and  when  x  >  —  4,  — ^  is  negative. 

dx2  dx2 

Thus  there  is  a  point  of  inflexion  at  (—  4,  2),  at  which  the  slope 
is  infinite,  and  the  bending  changes  from  the  positive  to  the  negative 
direction. 


Ex.  3.   Consider  the  curve 

dy  _ 


dx 


4  a:8, 


y  =  ar 

d2y 


dx2 


12  x2. 


d2y 


At  (0,  0),  ^  is  zero,  but  the 
dx2 

curve  has  no  inflexion,  for  — & 
dx2 
never  changes  sign  (Fig.  29). 


Fig.  29 


94  DIFFERENTIAL   CALCULUS 

51.  Concavity  and  convexity  toward  the  axis.  A  curve  is 
said  to  be  convex  or  concave  toward  a  line,  in  the  vicinity  of 
a  given  point  on  the  curve,  according  as  the  tangent  at  the 
point  does  or  does  not  lie  between  the  curve  and  the  line,  for 
a  finite  distance  on  each  side  of  the  point  of  contact. 


Fig.  30  a  Fig.  30  6 

First,  let  the  curve  be  convex  toward  the  #-axis,  as  in  the  left- 
hand  figure.     Then  if  y  is  positive,  the  bending   is   positive 

cl2v 
and  — "  is  positive :  but  if  y  is  negative,  the  bending  is  nega- 
dx2 

tive  and  —^  is  negative.     Hence  in  either  case  the  product 
dx2 

sin 

dx2 


y—^-  is  positive. 


Next,  let  the  curve  be  concave  toward  the  a>axis,  as  in  the 
right-hand  figure.     Then  if  y  is  positive,  the  bending  is  nega- 
tive and  — ^-  is  negative ;  but  if  y  is  negative,  the  bending  is 
dx2 

positive  and  — -^  is  positive.     Thus  in  either  case  the  product 
dx2 

d2v 
y — --  is  negative.     Hence  : 

dx2 

In  the  vicinity  of  a  given  point  (x,  y)  the  curve  is  convex  or 


concave  to  the  x-axis.  according  as  the  product  y  —^  is  positive  or 

dx2 


APPLICATIONS   TO   CURVE   TRACING  95 

EXERCISES 

1.  Examine  the  curve  y  —  2  —  3(x  —  2)5  for  points  of  inflexion. 

2.  Show  that  the  curve  a2y  =  x(a2  —  x2)  has  a  point  of  inflexion 
at  the  origin. 

3.  Find  the  points  of  inflexion  on  the  curve  y 


x2  +  4  a2 

■m 

4.  In  the  curve  ay  =  x«,  prove  that  the  origin  is  a  point  of  in- 
flexion if  m  and  n  are  positive  odd  integers. 

5.  Show  that  the  curve  y  =  c  sin  -  has  an  infinite   number   of 

a 

points  of  inflexion  lying  on  a  straight  line. 

6.  Show  that  the  curve  y(x2  -f-  a2)  =  x  has  three  points  of  inflexion 
lying  on  a  straight  line ;  find  the  equation  of  the  line. 

7.  If  y2  =f(xs)  is  the  equation  of  a  curve,  prove  that  the  abscissas 
of  its  points  of  inflexion  satisfy  the  equation 

[/'(*)]2  =  2/(z)  •/''(*). 

8.  Draw  the  part  of  the  curve  a'2y  =  ~ —  ax2  -f  2  a3  near  its  point 

o 

of  inflexion,  and  find  the  equation  of  the  stationary  tangent. 

9.  Show  that  the  curve  y  =  x2n  has  no  points  of  inflexion,  n  being 
any  positive  integer.     Sketch  the  curve. 

10.  Show  that  the  curve  (1  +  x2)y  =1  —  x  has  three  points  of  in- 
flexion, and  that  they  lie  in  a  straight  line. 

52.  Hyperbolic  and  parabolic  branches.  When  a  curve  has  a 
branch  extending  to  infinity,  the  tangents  drawn  at  successive 
points  of  this  branch  may  tend  to  coincide  with  a  definite  fixed 
line,  as  in  the  familiar  case  of  the  hyperbola.  On  the  other 
hand,  the  successive  tangents  may  move  farther  and  farther  out 
of  the  field,  as  in  the  case  of  the  parabola.  These  two  kinds 
of  infinite  branches  may  be  called  hyperbolic  and  parabolic. 

The  character  of  each  of  the  infinite  branches  of  a  curve  can 
always  be  determined  when  the  equation  of  the  curve  is  known. 


96  DIFFERENTIAL   CALCULUS 

53.  Definition  of  a  rectilinear  asymptote.  If  the  tangents  at 
successive  points  of  a  curve  approach  a  fixed  straight  line  as 
a  limiting  position  when  the  point  of  contact  moves  farther 
and  farther  along  any  infinite  branch  of  the  given  curve,  then 
the  fixed  line  is  called  an  asymptote  of  the  curve. 

This  definition  may  be  stated  more  briefly  but  less  precisely 
as  follows :  An  asymptote  to  a  curve  is  a  tangent  whose  point 
of  contact  is  at  infinity,  but  which  is  not  itself  entirely  at 
infinity. 

DETERMINATION  OF  ASYMPTOTES 

54.  Method  of  limiting  intercepts.  The  equation  of  the  tan- 
gent at  any   point  (x1}  yx)  being 


y-yi  =  ~(^-^i), 

ClXi 

the  intercepts  made  by  this  line  on  the  coordinate  axes  are 

dxl 

Xq  —  xx  —  yx  — 

dyx  J 

(1) 


Suppose  the  curve  has  a  branch  on  which  x  ==  oo  and  y  =  oo . 
Then  from  (1)  the  limits  can  be  found  to  which  the  intercepts 
x0,  y0  approach  as  the  coordinates  xlf  yx  of  the  point  of  contact 
tend  to  become  infinite.  If  these  limits  are  denoted  by  a,  b, 
the  equation  of  the  corresponding  asymptote  is 

a     b 

Except  in  special  cases  this  method  is  usually  too  compli-' 

cated  to   be   of   practical  use  in   determining  the   equations 

of  the  asymptotes  of  a  given  curve.      There  are  two  other 

methods,  which  together  will  always  suffice  to  determine  the 


APPLICATIONS   TO   CURVE   TRACING  97 

asymptotes  of  curves  whose  equations  involve  only  algebraic 
functions.  These  may  be  called  the  methods  of  inspection 
and  of  substitution. 

55.  Method  of  inspection.  Infinite  ordinates,  asymptotes  parallel 
to  axes.  When  an  algebraic  equation  in  two  coordinates  x  and 
y  is  rationalized,  cleared  of  fractions,  and  arranged  according 
to  powers  of  one  of  the  coordinates,  say  y,  it  takes  the  form 

a^+(6aj  +  c)r-1  +  (^  +  ^+/)r"2+  -  +«»-#  +  «»  =  <>, 

in  which  un  is  a  polynomial  of  the  degree  n  in  terms  of  the 
other  coordinate  x,  and  un.l  is  of  degree  n  —  1. 

When  any  value  is  given  to  x,  the  equation  determines  n 
values  for  y. 

Let  it  be  required  to  find  for  what  value  of  x  the  correspond- 
ing ordinate  y  has  an  infinite  value. 

For  this  purpose  the  following  theorem  from  algebra  will 
be  recalled : 

Given  an  algebraic  equation  of  degree  n, 

ayn  +  (3yn-1  +  yyn-2+  •••  =0; 

if  «  =  0,one  root  y  becomes  infinite;  if  a  =  0  and  /?  =  0,  two 
roots  y  become  infinite;  and  in  general  if  the  coefficients  of 
each  of  the  k  highest  powers  of  y  vanish,  the  equation  will 
have  k  infinite  roots. 

Suppose  at  first  that  the  term  in  yn  is  present ;  in  other 
words,  that  the  coefficient  a  is  not  zero.  Then,  when  any 
finite  value  is  given  to  x,  all  of  the  n  values  of  y  are  finite, 
and  there  are  accordingly  no  infinite  ordinates  for  finite 
values  of  the  abscissa. 

Next  suppose  that  a  is  zero,  and  b,  c,  not  zero.  In  this 
case  one  value  of  y  is  infinite  for  every  finite  value  of  x,  and 
el.  calc.  —  7 


98  DIFFERENTIAL  CALCULUS 

hence  the  curve  passes  through  the  point  at  infinity  on  the 

y  axis. 

There  is   one   particular  value  of  x,  namely,  x  =  ^—.  for 

b 

which  an  additional  root  of  the  equation  in  y  becomes  infinite. 
For,  when  x  has  this  value,  the  coefficient  bx  -f-  c  of  the  high- 
est power  of  y  remaining  in  the  equation  vanishes. 

Geometrically,  every  line  parallel  to  the  y  axis  has  one 
point  of  intersection  with  the  curve  at  infinity,  but  the  line 
bx-\-c  =  0  has  two  points  of  intersection  with  the  curve  at 
infinity.  A  line  having  two  coincident  points  of  intersection 
with  a  curve  is  a  tangent  to  the  curve ;  and  when  the  coinci- 
dent points  are  at  infinity,  but  the  line  itself  not  altogether  at 
infinity,  the  tangent  is  an  asymptote.  Hence,  an  ordinate  that 
becomes  infinite  for  a  definite  value  of  x  is  an  asymptote. 

Again,  if  not  only  a,  but  also  b  and  c  are  zero,  there  are 
two  values  of  x  that  make  y  infinite ;  namely,  those  values 
of  x  that  make  dx2  +  ex  +f=  0.  The  equations  of  the 
infinite  ordinates  are  found  by  factoring  this  last  equation; 
and  so  on. 

Similarly,  by  arranging  the  equation  of  the  curve  according 
to  powers  of  x,  we  can  easily  find  what  values  of  y  give  an 
infinite  value  to  x. 

Ex.  1.   In  the  curve 

2  xs  +  z2y  +  zy*  =  x2  —  y2  —  5, 

find  the  equation  of  the  infinite  ordinate,  and  determine  the  finite 
point  in  which  this  line  meets  the  curve. 

This  is  a  cubic  equation  in  which  the  coefficient  of  y8  is  zero. 

Arranged  in  powers  of  y  it  is 

y2(x  +  1)  +  yx2  +  (2  x8  t-  z2  +  5)  =0. 


APPLICATIONS  TO   CURVE   TRACING 


99 


When  x  =  —  1,  the  equation  for  y  becomes 

0  •  y2  +  y  +  2  =  0, 

the  two  roots  of  which  are  y  =  oo ,  y  =  —  2 ;  hence  the  equation  of 
the  infinite  ordinate  is  x  +  1  =  0.  The  infinite  ordinate  meets  the 
curve  again  in  the  finite  point  (—  1,  —  2). 

Since  the  tern^in  x8  is  present,  there  are  no  infinite  values  of  x 
for  finite  values  of  y. 

Ex.  2.   Show  that  the  lines  x  =  a,  and  y  =  0  are  asymptotes  to  the 
curve  a2x  =  y(x  —  a)2  (Fig.  31). 


Fig.  31 
Ex.  3.   Find  the  asymptotes  of  the  curve  x2(y  —  a)  +  xy2  =  a3. 

56.  Method  of  substitution.  Oblique  asymptotes.  The  as- 
ymptotes that  are  not  parallel  to  either  axis  can  be  found  by 
the  method  of  substitution,  which  is  applicable  to  all  algebraic 
curves,  and  is  of  especial  value  when  the  equation  is  given  in 
the  implicit  form 

/(*»y)  =  o.  (i) 


Consider  the  straight  line 

y  =  mx  +  b, 


(2) 


100  DIFFERENTIAL   CALCULUS 

and  let  it  be  required  to  determine  m  and  b  so  that  this  line 
shall  be  an  asymptote  to  the  curve  /(a*,  y)  =  0. 

Since  an  asymptote  is  the  limiting  position  of  a  line  that 
meets  the  curve  in  two  points  that  tend  to  coincide  at  infinity, 
then,  by  making  (1)  and  (2)  simultaneous,  the  resulting  equa- 

f(x,  mx  +  b)  =  0, 

is  to  have  two  of  its  roots  infinite.  This  requires  that  the 
coefficients  of  the  two  highest  powers  of  x  shall  vanish. 
These  coefficients,  equated  to  zero,  furnish  two  equations 
from  which  the  required  values  of  m  and  b  can  be  determined. 
These  values,  substituted  in  (2),  will  give  the  equation  of  an 
asymptote. 

Ex.  4.   Find  the  asymptotes  to  the  curve  y3  =  x2(2  a  —  x). 

In  the  first  place,  there  are  evidently  no  asymptotes  parallel  to 

either  of  the  coordinate  axes.     To  determine  the  oblique  asymptotes, 

make  the  equation  of  the  curve  simultaneous  with  y  =  mx  +  b,  and 

eliminate  y.     Then 

(mx  +  b)3  =  x2(2a-x), 

or,  arranged  in  powers  of  x, 

(1  +  m3)x3  +  (3  m2b  -  2  a)x2  +  3  b2mx  +  bs  =  0. 
Let  m3  +  1  =  0   and  3  m2b  -  2  a  =  0. 

Then  m  =  —  1,  />  =  =—; 

o 

hence  y  =  -  x  +  — 

o 

is  the  equation  of  an  asymptote. 

The  third  intersection  of  this  line  with  the  given  curve  is  found 

2  a 

from  the  equation  3  mb2x  -f  b8  =  0,  whence  x  =  —• 


APPLICATIONS  TO  CURVE'.  TRACING: 


161 


Fig.  32 

This  is  the  only  oblique  asymptote,  as  the  other  roots  of  the  equation 
for  m  are  imaginary. 

Ex.  5.    Find  the  asymptotes  to  the  curve  y(a2  +  x2)  =  a2(a  -  x). 

Y 


Fig.  33 

Here  the  line  y  =  0  is  a  horizontal  asymptote  by  Art.  55.     To  find 
the  oblique  asymptotes,  put  y  =  mx  +  b. 
Then  (mx  +  b)  (a2  4-  x2)  =  a2(a  -  x), 

i.e.  mx*  +  bx2  +  (ma2  +  a2)x  +  (a2b  -  a8)  =  0 ; 

hence  m  =  0,  b  =  0,     for  an  asymptote. 

Thus  the  only  asymptote  is  the  line  y  =  0  already  found. 


1#2  DIFFERENTIAL   CALCULUS 

57.  Number  of  asymptotes.  The  illustrations  of  the  last 
article  show  that  if  all  the  terms  are  present  in  the  general 
equation  of  an  nth  degree  curve,  then  the  equation  for  deter- 
mining m  is  of  the  nth  degree  and  there  are  accordingly  n 
values  of  ra,  real  or  imaginary.  The  equation  for  finding  b  is 
usually  of  the  first  degree,  but  for  certain  curves  one  or  more 
values  of  m  may  cause  the  coefficients  of  xn  and  xn~x  both  to 
vanish,  irrespective  of  b.  In  such  cases  any  line  whose  equa- 
tion is  of  the  form  y  =  myx  +  c  will  have  two  points  at  infinity 
on  the  curve  independent  of  c ;  but  by  equating  the  coefficient 
of  xn~2  to  zero,  two  values  of  b  can  be  found  such  that  the  re- 
sulting lines  have  three  points  at  infinity  in  common  with  the 
curve.  These  two  lines  are  parallel ;  and  it  will  be  seen  that 
in  each  case  in  which  this  happens  the  equation  defining  m 
has  a  double  root,  so  that  the  total  number  of  asymptotes  is 
not  increased.  Hence  the  total  number  of  asymptotes,  real 
and  imaginary,  is  in  general  equal  to  the  degree  of  the  equation 
of  the  curve. 

This  number  must  be  reduced  whenever  a  curve  has  a  para- 
bolic branch,  since  in  this  case  a  value  of  m  which  makes  the 
coefficient  of  xn  vanish  does  not  correspond  to  any  finite  value 
of  b. 

Ex.  6.  Find  the  asymptotes  of  the  curve  (x  —  y) 3  =  2  x.  The 
equation  in  m  is  (m  —  l)3  =  0.  The  coefficient  of  z3  vanishes  identi- 
cally when  m  =  1 ;  that  of  x  is  3(?n  —  l)b2  -  2  which  cannot  be  made 
to  vanish  for  any  finite  value  of  b  when  m  =  1*  The  curve  has  no 
asymptotes. 

Ex.  7.     Find  the  asymptotes  of  the  curve 
0r-l)(2-^ 
*   '  x  -  3 

and  trace  the  curve.     (Fig.  34.) 


APPLICATIONS  TO   CURVE   TRACING 


103 


Fig.  U 


EXERCISES 

Find  the  asymptotes  of  each  of  the  following  curves: 


1.   y(a2  -  x2)  =  b(2z  +  c). 


9  x2-2ax 


3.    x2y2  =  a\x2  -  y2). 

b* 


4.    y  =  a  + 


5.  yz  =  x2(a  —  x). 

6.  y\x  -  l)  =  x2. 

15.    x8  +  2  x2y  -  xy2  -  2  ys  +  4  y2  +  2  xy  +  y  =  1 


7.  (*  +  a)y«=(y+6)*«. 

8.  x2y2  =  a:3  -f  #  +  y. 

9.  xy2  +  x2y  =  a3. 

10.  y(x2  +  3a2)=a* 

11.  x3  -  3  axy  +  y8  =  0. 

12.  j;3  +  y3  =  a3- 

13.  x4  -  x2y2  +  a2x2  +  b*  =  0. 

14.  x4  -  y4  -  a2*^. 


104 


DIFFERENTIAL   CALCULUS 


POLAR   COORDINATES 

58.  When  the  equation  of  a  curve  is  expressed  in  polar 
coordinates,  the  vectorial  angle  0  is  usually  regarded  as  the 
independent  variable.  To  determine  the  direction  of  the 
curve  at  any  point,  it  is  most  convenient  to  make  use  of  the 
angle  between  the  tangent  and  the  radius  vector  to  the  point 

of  tangency. 

Let  P,  Q  be  two  points  on  the 
curve  (Fig.  35).  Join  P,  Q  with 
the  pole  0,  and  drop  a  perpendic- 
ular PM  from  P  on  OQ.  Let  p, 
6  be  the  coordinates  of  P ;  p  +  Ap, 
0  +  A6  those  of  Q.  Then  the  angle 
POQ  =  A0',  PM=p  sin  A0;  and 
Fig.  35  MQ  =  OQ-  OM=p  +  Ap-p  cos  A0. 


Hence 


tan  MQP 


p  sin  A6 


p  -f  Ap  —  p  cos  A0 


When  Q  moves  to  coincidence  with  P,  the  angle  MQP 
approaches  as  a  limit  the  angle  between  the  radius  vector 
and  the  tangent  line  at  the  point  P.  This  angle  will  be 
designated  by  if/. 


Thus 


tan  if/ 


lim  * 

A0  =  O 


p  sin  A0 


p  +  Ap  —  p  cos  A0 


But  p  (1  -  cos  A0)  =  2P  sin2  i  A0, 


hence  tan  i/>  =  A0l™  0 


p  sin  A0 


1  A/i     sin  i  A0  .  Ap 

smiA0--r-hr-  +  ^ 


±A0 


APPLICATIONS   TO   CURVE   TRACING 


105 


Since  Allr?  ft  sm  -        ==  1,  the  preceding  equation  reduces  to 


i±0 


tan  if/  = 


dp 

dB 


dO 
dp 


(3) 


Ex.  1.  A  point  describes  a  circle  of  radius  p. 
Prove  that  at  any  instant  the  arc  velocity  is  p  times 
the  angle  velocity, 


ds 
dt 


'  dt  ' 


dt' 


Ex.  2.     When  a  point  describes  a  given 
curve,  prove  that  at  any  instant  the  velocity 

—  has  a  radius  component  l-&-  and  a  com- 
dt  dt 

ponent  perpendicular  to  the  radius  vector 

p  —  ,  and  hence  that 
r  dt 


Fig.  37 


cos  \p 


ds 


i  dO    ,         .  r< 

,  sin  \p  =  p  — ,  tan  \p  =  p  - 


ds 


This  furnishes  a  dynamical  proof  of  equation  (3). 

If 


59.    Relation  between  -£-  and  p  V 
doc  dp 


the  initial  line  is  taken  as  the  axis  of  x, 
the  tangent  line  at  P  makes  an  angle  $ 
with  this  line. 

Hence  0  -f  \p  =  <£ ; 

dd\      ..   _Udyy 


i.e.,  0  +  tan" 


p  —  i  =  tan" 
dPj 


dx 


Fig.  38 


60.  Length  of  tangent,  normal,  polar  subtangent,  and  polar  sub- 
normal. The  portions  of  the  tangent  and  normal  intercepted 
between  the  point  of  tangency  Pand  the  line  through  the  pole 
perpendicular  to  the  radius  vector    OP,  are  called  the  polar 


106 


DIFFERENTIAL  CALCULUS 


tangent  length  and  the  polar  normal  length;  their  projections  on 
this  perpendicular  are  called  the  polar  subtangent  and  polar 
subnormal. 


Fig.  39  a 


Fig.  39  b 


Thus,  let  the  tangent  and  normal  at  P  (Figs.  39  a,  b)  meet  the 
perpendicular  to  OP  in  the  points  jVand  M.     Then 

PN  is  the  polar  tangent  length, 
PM  is  the  polar  normal  length, 
ON  is  the  polar  subtangent, 
OM  is  the  polar  subnormal. 

They  are  all  seen  to  be  independent  of  the  direction  of  the 
initial  line.    The  lengths  of  these  lines  will  now  be  determined. 


Since  PN=OP-  sec  OPN=  p  sec  ^ 

flO 


4 


•\|j+1 


hence      polar  tangent  length  =  p 


dp 

(16 
dP 


\P 


'+D 


v 


'<$)'■ 


Again,  ON=  OP  tan  OPN=  p  tan  if/  =  p 


hence 


polar  subtangent  =  p'2— 
dp 


2d$ 
dp' 


APPLICATIONS   TO   CURVE   TRACING  107 


PM  =  OP  •  esc  OPN=  p  esc  ^  =  yj/  +  (&\  , 

hence     polar  normal  length  =  -v/p2 -h  (  — ^* )  • 

OM=  OP  cot  OPN  =  *£, 
dO 

hence  polar  subnormal  =  -£  • 

The  signs  of  the   polar   tangent  length  and  polar   normal 

length  are  ambiguous  on  account  of  the  radical.      The  direc- 

d6 
tion  of   the  subtangent   is   determined  by  the    sign  of  p2 —  • 

09.  dp 

When  —  is  positive,  the  distance  ON  should  be  measured  to 
dp 

the  right,  and  when  negative,  to  the  left  of  an  observer  placed 

at    0  and   looking    along    OP;    for   when    0    increases   with 

d6 
p,  —  is  positive  (Art.  28),  and  \p  is  an  acute  angle   (as   in 
dp 

d6 
Fig.  39  b) ;  when  6  decreases  as  p  increases,   _   is  negative, 

and  if/  is  obtuse  (Fig.  39  a). 


EXERCISES 

1.  In  the  curve  p  =  a  sin  0,  find  \J/. 

2.  In  the  spiral  of  Archimedes  p  =  aO,  show  that  tan  xp  =  6  and 
find  the  polar  subtangent,  polar  normal,  and  polar  subnormal.  Trace 
the  curve. 

3.  Find  for  the  curve  p2  =  a2  cos  2  0  the  values  of  all  the  expres- 
sions treated  in  this  article. 

4.  Show  that  in  the  curve  p6  =  a  the  polar  subtangent  is  of  con- 
stant length.     Trace  the  curve. 

5.  In  the  curve  p  =  a(l  —  cos  0),  find  if/  and  the  polar  subtangent. 


108  DIFFERENTIAL  CALCULUS 

6.  Show  that  in  the  curve  p  =  b  •  e9cota  the  tangent  makes  a  con- 
stant angle  a  with  the  radius  vector.  For  this  reason,  this  curve  is 
called  the  equiangular  spiral. 

7.  Find  the  angle  of  intersection  of  the  curves 

p  =  a{l+  cos  0),  p  =  h(l  —  cos  6). 

a 

8.  In  the  parabola  p  =  a  sec2-,  show  that  cp  +  $  =  tt. 

EXERCISES  ON   CHAPTER  VII 

Trace  the  following  curves.  Find  asymptotes,  intervals  of  in- 
creasing and  decreasing  ordinate  and  direction  of  bending,  as  well  as 
intercepts  on  the  axes. 

1.  y  =  xs  +  2  x2  -  7  x  +  1.  5.   y2  =  xs. 

2.  y2  =  x3  +  2  x2  -7  x  +  1.  6.   ay2  =  x3  -  hx2. 

3.  y  =  02-l)2.  7.   z4-7/4  =  2x. 

4.  x3  +  ?/3  =  1. 

In  the  following  curves  find  \p,  determine  whether  p  can  become 
infinite,  and  obtain  the  (angular)  intervals  of  increasing  and  decreas- 
ing p. 

8.  p  =  a  cos  2  6.  10.   p  =  «(1  -  cos  0). 

a 

9.  p  =  a  sin  3  0.  "■•  p  =  «sec2-. 


CHAPTER  VIII 

DIFFERENTIATION   OF   FUNCTIONS    OF   TWO   VARIABLES 

Thus  far  only  functions  of  a  single  variable  have  been  con- 
sidered. The  present  chapter  will  be  devoted  to  the  study  of 
functions  of  two  independent  variables  x,  y.  They  will  be 
represented  by  the  symbol 

2  =/(?,  y)- 

If  the  simultaneous  values  of  the  three  variables  x,  ?/,  z  are 
represented  as  the  rectangular  coordinates  of  a  point  in  space, 
the  locus  of  all  such  points  is  a  surface  having  the  equation 

61.  Definition  of  continuity.  A  function  z  of  x  and  y, 
z=f(x,  ?/),  is  said  to  be  continuous  in  the  vicinity  of  any  point 
(a,  b)  when  /(a,  b)  is  real,  finite,  and  determinate,  and  such 

hYTof(a  +  h,b  +  k)=f(a,b), 
k  =  0 

however  h  and  k  approach  zero. 

When  a  pair  of  values  a,  b  exists  at  which  any  one  of  these 
properties  does  not  hold,  the  function  is  said  to  be  discontinu- 
ous at  the  point  (a,  b). 

E.g.,  let  z=x-+JL. 

x  -  y 

When  x  =  0,  then  z  —  —  1  for  every  value  of  y ;  when  y  =  0  then 
2=4-1  for  every  value  of  x.     In  general,  if  y  —  mx, 

-  -  1  +  m 
1  —  m 

109 


110  DIFFERENTIAL   CALCULUS 

and  z  may  be  made  to  have  any  value  whatever  at  (0,  0)  by  giving  an 
appropriate  value  to  m. 

Geometrically  speaking,  when  the  point  (x,  y)  moves  up  to  (G,  0), 
the  limiting  value  of  the  ordinate  z  depends  upon  the  direction  of 
approach. 

62.    Partial  differentiation.     If  in  the  function 

*  =/to  y) 

a  fixed  value  yx  is  given  to  y,  then 

is  a  function  of  x  only,  and  the  rate  of  change  in  z  caused  by  a 
change  in  x  is  expressed  by 

dz  = — dx,  (1) 

dx 

dz 
in  which  —  is  obtained  on  the  supposition  that  y  is  constant. 
dx 

To  indicate  this  fact  without  the  qualifying  verbal  state- 
ment, equation  (1)  will  be  written  in  the  form 

d^Ax.  (2) 

OX 

dz 
The  symbol  —  represents  the  result  obtained  by  differentiat- 
ed 

ing  z  with  regard  to  x,  the  variable  y  being  treated  as  a  con- 
stant; it  is  called  the  partial  derivative  of  z  with  regard  to  x. 

From  the  definition  of  differentiation,  Art.  6,  the  partial 
derivative  is  the  result  of  the  indicated  operation 

dz=    lim    f(x  +  Ax,  y)-f(x,y) 
dx      Ax~°  Ax 

Similarly,  the  symbol  —  represents  the   result   obtained  by 

differentiating  z  with  regard  to  y,  the  variable  x  being  treated 


FUNCTIONS  OF  TWO   VARIABLES  111 

as  a  constant;   it  is  called  the  partial  derivative  of  z  with 
regard  to  y. 

The  partial  derivative  of  z  with  regard  to  y  is  accordingly  the 
result  of  the  indicated  operation 

dz  =    lim    f(x,  y  +  &y)-f(x,y) 

dy      Ay  =  °  Ay 

dz 
dxz  =  — die  is  called  the  partial  x-differential  of  z,  and 
dx 

dz 
cLz  =  — dy  is  called  the  partial  y-differential  of  z. 
by 

EXERCISES 

1.  Given  u  =  x*  +  3 x*y*  -  7 xys,  prove  that  x~-hy^=iu. 

ox        By 

2.  Given   u  =  tan-1^,  show  that  x—  +  y  —  =  0. 

x  dx        dy 

3.  u  =  log  (e*  +  e»)  ;  find  ^  +  & 

dx     dy 

4.  n  =  sin  ary;  find  ^  +  ^. 

dx      dy 

5.  u  =  log  (x  +  V^T^)  ;  find  x~  +  y|^ 

6.  m  =  log  (tan  x  +  tan  y  +  tan  2) ;  show  that 

sin  2x^  +  sin  2  y^  +  sin  2 *<?H  =  2. 
dx  dy  dz 

7.  II  =  log  (a;  +  y)  ;   show  that  f?  +  ^  =  1. 

5a:      dy      ew 

8.  ti  =  -3L;  show  that  *^ +  «£!?=«. 

*+3/  dx      *dy 

9.  u  =  (y-«)(2-x)(x-y);  show  that  ^  +  f^  +  ^  =  0. 

dx      dy      dz 

10-  »  =  ^  show  that  g+g  =  (»  +  »-D«. 

11.   u  =  log  (x3  +  y8  +  z8  -  3  xyz)  ;  show  that 
du  ,   du  .Qu  3 


dx       dy      dz      x  +  y-\-  z 


112  DIFFERENTIAL   CALCULUS 

63.  Total  differential.  If  both  x  and  y  are  allowed  to  vary 
in  the  function  z=f(x,  y),  the  first  question  that  naturally 
arises  is  with  regard  to  the  meaning  of  the  differential  of  z. 

Let  zx  =  f(xx,%), 

and  zx  +  Az  =  f(xl-\-  Ax,  yx-\-  Ay) 

be  two  values  of  the  function  corresponding  to  the  two  pairs 
of  values  of  the  variables  a^,  yx  and  xx  +  Ax,  yx  +  Ay. 
The  difference 

Az  =  f(xx  +  Ax,  yx  +  Ay)  -  f(xx,  yx) 

may  be  regarded  as  composed  of  two  parts,  the  first  part  being 
the  increment  which  z  takes  when  x  changes  from  xx  to  xx  +  Ax, 
while  y  remains  constant  (yz=yx),  and  the  second  part  being 
the  additional  increment  which  z  takes  when  y  changes  from 
2/i  to  yx  -f-  Ay,  while  x  remains  constant  (x  =  xx  +  Ax).  The 
increment  Az  may  then  be  written 

Az  =  f{xx  +  Ax,  ^  +  Ay)  -  ffa  +  Ax,  ^) 

4-/(«i  +  Ax,  2/0  -f(xx,  yx) 

-  /fa  +  Aa?>  2/1  +  A-y)  ^  /fa  +  A:g>  yQ  a?/ 

Ay 

+  f(xx  +  Ax,yx)-f(xx,yl)  ^x 
Ax 

From  the  theorem  of  mean  value,  Art.  39,  the  last  equation 
may  be  written 

Az  =  A/fa  +  6 Ax,  yx)  Ax  +  A/fa  +  Ax,  y,  +  $xAy)  Ay.    (3) 
dx  dy 

It  represents  the  actual  increment  Az  which  the  dependent 
variable  z  takes  when  the  independent  variables  x  and  y  take 
the  increments  Ax  and  Ay. 


FUNCTIONS   OF   TWO   VARIABLES 


113 


To  illustrate,  let  z  =  f(x,  y)  be  the  equation  of  a  surface  (Fig.  40). 
Let  Ai  =  (xi,  y\),  A2  =  (xi  +  Ax,  y\),  A%=  (xi  +  Ax,  y\  +  Ay),  so  that 
AiPi=f(xh  yx),  AiPi~f{xi  +  Ax,  yx),  AzPi=f{xl  +  Ax,  yx  +  Ay), 
Q2P2=/(2i  +  Ax,  yi)  -f(xh  yi)  =  Aiz, 
QzPz=f(xi  +  Ax,  yi  +  Ay)  -f(xi  +  Ax,  y{)  =  A2z, 
RzP&=f{xi  +  Ax,  yx  +  Ay)  -/(a*,  yi)  =  Axz  +  A2z  =  Az. 

As  the  moving  point  P  passes  from  P  to 
P-2  along  the  plane  curve  P\P2,  the  ordinate 
takes  the  increment 


Tt 


n 

i? 

f-'  Q3 

' 

Q* 

/ 

i?a 

A 

' 

V 

A2 

A3 
Fig.  40 


where  the  derivative  is  taken  at  the  inter- 
mediate point  x  =  x\  +  6 Ax,  y  —  y\  (Art.  39). 
Similarly,  as  P  passes  from  P2  to  Pz  along    / 
the  plane  curve  P2Ps,  the  ordinate  takes  the 
further  increment 

where  the  derivative  is  taken  at  the  intermediate  point  y  =  y\  4-  diAy, 
x  =  xi  -f  Ax. 

The  sum  of  these  two  partial  increments  gives  the  total  increment  Az. 

In  the  preceding  equation  (3)  let  Ax,  Ay,  Az  be  replaced  by 
c  •  dx,  €  •  dy,  e  •  dz  respectively,  in  which  dx,  dy  are  entirely 
arbitrary.  After  removing  the  common  factor  e,  let  e  approach 
zero.     The  result  is 

(4) 


dx  dy 


The  differential  dz  denned  by  this  equation  is  called  the  total 
differential  of  z.  It  is  not  an  actual  increment  of  z,  but  the 
increment  which  z  would  take  if  its  change  continued  uniform 
while  x  changed  from  xx  to  xx  +  dx  and  y  changed  from  yx  to 
Vi  +  dy. 

EL.    CALC  — 8 


114  DIFFERENTIAL   CALCULUS 

In  other  words,  dz  is  the  rate  of  change  of  the  variable  2 
when  the  independent  variables  x  and  y  change  simultane- 
ously at  the  rates  of  dx,  dy  respectively. 

Equation  (4)  may  be  written  in  the  form 

dz  =  ^dx  +  -^dy  =  dxz  +  dyz, 
dx  dy 

from  which  the  following  theorem  can  be  stated ;  the  total  dif- 
ferent icd  of  a  function  of  two  variables  is  equal  to  the  sum  of  iU 
partial  differentials  taken  with  regard  to  the  separate  variables, 
or  the  total  rate  of  change  of  z  is  equal  to  the  sum  of  its  par- 
tial rates. 

The  same  method  can  be  applied  directly  to  functions  of 
three  or  more  variables.  Thus,  if  u  is  a  function  of  the  vari- 
ables x,  y,  z,  u=^(x,yJz)) 

then  du  =  ~dx-\-^-  dy  +  — *  dz. 

ox  dy  dz 

Ex.1.    Given  z  =  axy'2  -\-bx2y  +  ex3  +  cy, 

then  dz  =  (ay2  +  2  bxy  +  3  cx2)dx  +  (2  axy  +  bx'2  +  c)dy. 

Ex.  2.    Given  u  =  tan-1--,  show  that  du  —     ^  ~  "  — 
x  x'1  +  y'2 

Ex.  3.  Assuming  the  characteristic  equation  of  a  perfect  gas, 
vp  =  Rt,  in  which  v  is  volume,  p  pressure,  t  ahsolute  temperature,  and 
11  a  constant,  express  each  of  the  differentials  dv,  dp,  dt,  in  terms  of 
the  other  two. 

Ex.  4.  A  particle  moves  on  the  spherical  surface  x2  +  y2  +  z2  =  a2 
in  a  vertical  meridian  plane  inclined  at  an  angle  of  60°  to  the  em- 
plane. If  the  a; -component  of  its  velocity  is  ^-  feet  per  second  when 
x  —  -  ,  find  the  y-component  and  the  ^-component  velocities. 

Since                           z  =  Va2  —  x2  —  y2, 
then  dz  = ^ ■ y-V 


Va'2  —  x2  —  y'2      vV-  —  x2  —  y'2 


FUNCTIONS   OF   TWO    VARIABLES  115 

But  since  dx  =  — ,  and  the  equation  of  the  given  meridian  plane  is 

V  =  x  tan  60°,  hence  dy  —  VS  dx  =  a  — ,  and  y  =  ^——  •     Therefore 
u  j  10  4 

dz  =  -  —  -  &  =  -  —  in  feet  per  second. 
2V3       2  15 

Ex.  5.  A  triangle  has  a  base  of  10  units  and  an  altitude  of  6  units. 
The  base  is  made  to  increase  at  the  rate  of  2  units  and  the  altitude 
to  decrease  at  the  rate  of  \  unit.     At  what  rate  does  the  area  change? 

Ex.  6.    A  point  on  the  hvperboloid  x2  —  &- =  1  in  the  position 

4      5 

x  =  2,  y  —  2  moves  so  that  x  increases  at  the  rate  of  2  units  per  sec- 
ond, while  y  decreases  at  the  rate  of  3  units  per  second.  Find  the 
rate  of  change  of  z. 

Ex.  7.  If  the  area  of  a  rectangle  A  =  xy  is  incorrectly  measured 
owing  to  a  small  error  dx,  dy  in  the  length  of  each  side,  how  close 
is  dA  =  xdy  +  ydx  to  the  actual  error  in  the  area? 

64.  Total  derivative.  If  in  the  relation  z=f(x,  y),  the  vari- 
ables x,  y  are  not  independent,  but  both  are  functions  of 
another  variable  s,  the  process  of  the  preceding  article  can 
still  be  applied.  The  variable  z  is  now  a  function  of  s,  and 
its  derivative  as  to  s  may  be  expressed  in  the  form 

dz  _  dz  dx      dz  dy 
ds      dx  ds      dy  ds 

In  particular,  if  y  is  not  independent,  but  is  a  function  of 

x,  then  s  may  be  chosen  as  x  itself,  and  the  preceding  equation 

becomes  ' 

dz  _  dz      dz  dy 

dx      dx      dy  dx 
If  the  functional  relation  between  x  and  y  is  given, 

y  =  <*>0)> 


116  DIFFERENTIAL   CALCULUS 

dz 
then  the  same  result  will  be  obtained,  whether   —  is  deter- 

dx 

mined  by  the  present  method,  or  y  is  first  eliminated  from  the 
relation 

*=f(®,  y), 

and  the  resulting  equation  is  differentiated  as  to  x.  The 
method  of  this  article  frequently  shortens  the  process. 

It  is  here  well  to  note  the  difference  between  —  and  —  • 

dx  dx 

The  former  is  the  partial  derivative  of  the  functional  expres- 
sion for  z  with  regard  to  x,  on  the  supposition  that  y  is  con- 
stant. The  latter  is  the  total  derivative  of  z  with  regard  to 
x,  when  account  is  taken  of  the  fact  that  y  is  itself  a  func- 
tion of  x. 

In  the  former  case  the  differentiation  with  regard  to  x  is 
merely  explicit ;  in  the  latter  it  is  both  explicit  and  implicit. 

dz 


Ex.  1.    Given  z  =  Vx2  -f  y2,  y  =  log  x;  find 

dx 

dz  _         x  y         dy 

dx      y/x2  _f.  yi      \Ac2  +  y2  dx 


d]L-_ 
dx 

_1 

x 

dz 

X2 

+  y 

dx 

xyJx 

2    +    yl 

hence 


Ex.  2.    If  z  =  tan-i  ^-  and  4  x2+  y2  =  1,  show  that  ^  =  — . 
2x  dx        y 

65.    Differentiation  of  implicit  functions.     If,  in  the  relation 
z  =f(x,  y),  z  is  assumed  to  be  constant,  then 

dz  =  0 : 


FUNCTIONS  OF  TWO  VARIABLES  117 

hence  ¥-dx +  dfdy  =  0,  (1) 

dx  By 


from  which  ^  =  -37-  (2) 


dy_ 
dx 

df 
dx 

dy 

In  all  such  cases  either  variable  is  an  implicit  function  of 
the  other,  and  thus  the  last  equation  furnishes  a  rule  for 
finding  the  derivative  of  an  implicit  function. 

Ex.  1.    Given  x3  +  y3  +  3  axy  =  c,  find  -^. 

dx 

Since  (3  x2  +  3  ay)  +  (3  y2  +  3  ax)  ^  =  0,  ^  =  -  x*  +  ay . 

dx  dx  y2  +  ax 

Ex.  2.  /(a*  +  ty)  =  c ;    J£  =  a/'(ax  +  by)  ;     |£  =  &/'(«*  +  by) ; 

<7y  _  _  a 
</x~      &' 

Ex.  3.    If  ax2  +  2  fary  4-  %2  +  2  gx +  2fy  +  c  =  0,  find  ^. 

</x 

Ex.  4.   Given  x4  -  y4  =  c,  find^. 

</x 

Ex.  5.  If  x  increases  at  the  rate  of  2  inches  per  second  as  it  passes 
through  the  value  x  =  3  inches,  at  what  rate  must  y  change  when 
y  =  1  inch  in  order  that  the  function  2  xy2  —  3  x2y  shall  remain 
constant  ? 


If 

u  =  2  xy2  -  3  x2y, 

then 

du        n     i        a              du         a 

—  =  2y2-Qxy,    ^-=±Xy- 
dx                        dy 

du                                 dy 

dy  _      dx  _      2y2-6xy  _dt 
dx  ~      du  ~      4  xy  —  3  x2  ~~  </x 

dy                             dt 

Since  x  =  3,  y  —  1,  -y  =  2,  hence,  -y  =  -  2T25  inches  per  second. 


118  DIFFERENTIAL   CALCULUS 


Ex. 

6. 

u  —  v2  -f  vy,  v  =  log  s,  y  =  e8.    Find  — 

Ex. 

7. 

u  =  sin"1  (r  -  a),  r  =  3  7,  s  =  4  *3.     Fin 

Ex. 

8. 

e»  _  c*  +  a^  =  o.     Find  ^. 

Ex. 

9. 

sin  (*#)  -  e**  -  xhj  =  0.     Find  ^. 

du 
dt 


It  is  to  be  noticed  that  the  result  of  differentiating  any  implicit 
function  of  x,  y  by  the  method  of  the  present  article  will  agree  with 
the  result  of  differentiation  according  to  the  rules  of  Chapter  II. 

66.    Geometric   interpretation.      Geometrically,  the   equation 

z  =f(x,  y)  represents  a  surface.     The  equation  y  =  yx  defines 

a  plane  parallel  to  the  ^-coordinate  plane.     The  two  equations 

treated  simultaneously  therefore  define  the  plane  section  made 

on  the  surface  z  =  f(x,  y)  by  the  plane  y  =  yx.     The  derivative 

dz 

— -  defines  the  slope  of  the  tangent  line  to  this  curve  at  the 

axx 

point  (xl}  yu  zx). 

Similarly,  the  plane   x  =  xl  cuts  the  surface  in  a  section 

parallel  to  the  yz-coordinate  plane.     The  slope  of  the  tangent 

dz 
line  to  this  second  curve  is  defined  by  — -.     The  equations  of 

these  two  lines  are 

V  =  2/i,  z  -  z1  =  — i  (x  -  xx), 
dxx 

dz 
x  =  xx,  z-zx  =  -±  (y  -  ?/,). 

They  have  the  point  (xx,  yx,  zx)  in  common ;  hence  the  two  lines 
will  define  a  plane.  The  equation  of  any  plane  through  the 
first  line  will  be  of  the  form 


dXi 


+  *(//-*/,)=  o, 


FUNCTIONS   OF   TWO   VARIABLES  119 

and  similarly,  the  equation  of  any  plane  through  the  second 
line  will  be  of  the  form 


oyi 


+  k'(x-x1)=0. 


These  two  equations  will  be  identical  when 

dz,      ,  dz-, 

K  =  ~  ^H~>  k  =  —  — -j 
tyi  Mi 


hence  the  equation  of  the  plane  containing  both  lines  is 

2  -  *i  =  -p-  (x  -  a^H  ^-{y  -  vi). 

dxx  dyx 

It  is  called  the  tangent  plane  to  the  surface  z  =  f(x,  y)  at  the 
point  («!,  yv  zx). 
From  the  equation 

dM  =  %Ldx+pLdg,  (3) 

axl  dy1 

it  is  seen  that  if  x,  y  receive  the  arbitrary  increments  dx,  dy, 
then  the  increment  dz  is  defined  by  the  sums  of  the  products 
of  these  increments  by  the  corresponding  partial  derivatives. 
Thus,  if  dx  =  x  —  X],  dy  =  y  —  yl9  dz  =  z  —  zl9  it  is  seen  that 
the  point  (x,  y,  z)  always  lies  in  the  tangent  plane  to  the  sur- 
face z  =f(x,  y),  however  the  increments  dx  and  dy  approach 
zero. 

Moreover,  the   equations  of   the  line   joining    (xlf  yl9  zx)  to 
Xj  -f  A#,  ?/!  +  Ay,  zy  +  Az  on  the  surface  will  be  of  the  form 

x  -  :>\  __  y  -  yl  _  z  -  zx 
Ax  A?/  Az 


120  DIFFERENTIAL   CALCULUS 

Now  as  Ax,  Ay  approach  zero,  the  point  always  remaining 

on  the  surface,  the  line  becomes  a  tangent  in  the  limit,  and  its 

equations  are 

x~x1  =  y-y1  =  z-zx  (i) 

dx  dy  dz    '  W 

wherein  dx,  dy  depend  upon  the  direction  of  approach,  and  dz 
is  defined  by  (3). 

But  a  tangent  line  to  the  surface  is  also  tangent  to  any  plane 
section  passing  through  the  line,  and  the  line  (4)  is  seen  to  lie 
in  the  tangent  plane,  hence : 

TJie  tangent  lines  to  all  the  plane  sections  of  the  surface 
z  =f(x,  y)  passing  through  the  point  (xx,  yx,  Zj)  lie  in  the  tangent 
plane  at  that  point. 

The  line  through  (xY,  yly  z±)  perpendicular  to  the  tangent 

plane  dz  dz 

t-^^-^+dy1^-^ 

is  called  the  normal  to  the  surface  at  the  point  (x^  yx,  z^).     Its 

equations  are 

X-Xl      y-yx.    z-zx 


dzx  dzx  —  1 

dxx  dyx 

If  the  equation  of  the  surface  is  given  in  the  implicit  form 

F(x,  y,  z)  =  0,  then  since 

dF,     ,  dF ,     ,  dF,       ft 

-—dx-\ rt?y  -| dz  =  0, 

ox  dy  '        dz 

the  equation  of  the  tangent  plane  becomes,  if  F(a;1,  ylf  z^)  =  F1? 
dxt  dt/t  dzl 


FUNCTIONS   OF  TWO   VARIABLES  121 

and  those  of  the  normal  are 

3-3*1  =  y—V\  _  g-gi 

bFy         dF\         BFX  ' 
dxx  dy1  dzj 

EXERCISES 

1.  Show  that  the  plane  z  =  0  touches  the  surface  z  =  xy  at  (0,  0,  0). 

2.  Find  the  equation  of  the  tangent  plane  to  the  paraboloid 
z  =  2  x  2  +  4  y2  at  the  point  (2,  1,  12). 

3.  Find  the  equations  of  the  normal  to  the  hyperboloid 

x2_4y2+2za=6at  (2,2,3). 

4.  Show  that  the  normal  at  any  point  (xh  yi,  z{)  on  the  sphere 

x2  +  y2  +  z2  =  16  will  pass  through  the  center. 

5.  Find  the  equation  of-the  tangent  plane  at  any  point  (x\,  yh  z\) 
of  the  surface  x1  +  y*  +  z5  =  a*  and  show  that  the  sum  of  the  squares 
of  the  intercepts  which  it  makes  on  the  coordinate  axes  is  constant. 

6.  Show  that  the  volume  of  the  tetrahedron  cut  from  the  coor- 
dinate planes  by  any  tangent  plane  to  the  surface  xyz  =  a3  is  constant. 

7.  The  sphere  x2  +  y2  +  z2  =  14  and  the  ellipsoid  3  x2  +2  y2  +  z2  =  20 
pass  through  the  point  (—1,-2,-3).  Determine  the  angle  at 
which  their  tangent  planes  at  this  point  intersect. 

8.  How  far  distant  from  the  origin  is  the  tangent  plane  to  the 
ellipsoid  x2  +  3  y2  +  2  z2  =  9  at  the  point  (2,  -1,  1)  ? 

9.  Find  the  equation  of  the  tangent  plane  and  of  the  normal  to 
the  cone  z2  =  2  x2  +  y2  at  (xly  yi,  z\)  on  the  surface.  Show  that  the 
plane  will  always  pass  through  the  vertex  of  the  cone. 

10.  Find  the  equations  of  the  tangent  line  to  the  circle 

x2  +  y2  +  z2  =  25, 

x  +  z  =  5, 
at  the  point  (2,  2V3,  3). 


122  DIFFERENTIAL   CALCULUS 

67.  Successive     partial     differentiation.       The     expressions 

dz    dz 

— ,  —  which  were   denned  in  Art.  62  are  functions   of   both 

ox  oy 

x  and  y. 

dz 
If  —  is  differentiated  partially  as  to  x,  the  result  is  written 
ox 

d_fdz\==dhm 
dx\dxj      dx2 

This  expression  is  called  the    second   partial  derivative  of 
z  as  to  x. 

Similarly,  the  results  of  the  operations  indicated  by 

JL{<>?\  JL(te\  JL(te\ 

dy\dxf  dx\dyf  dy\dyj 

dh       dh     d2z 
are  written  -— -,  — — ,  -—   respectively. 
oy  ox  ox  oy  oyz 

Beginning   with   the   left,   we   call   these   expressions    the 

second  partial  derivative  of  z  as  to  x  and  y,  the  second  partial 

derivative  of  z  as  to  y  and  x,  and  the  second  partial  derivative 

of  z  as  to  y. 

68.  Order  of  differentiation  indifferent. 
Theorem.     The  successive  partial  derivatives 

d2z       dh 
By  dx  dx  dy 

are  equal  for  any  values  of  x  and  y  in  the  vicinity  of  which 
z  and  its  first  and  second  partial  x-  and  ?/-derivatives  are 
continuous. 

The  truth  of  this  theorem  will  be  assumed.     Tt  should  be 
verified  for  special  cases  as  in  the  following  examples. 


FUNCTIONS   OF   TWO   VARIABLES  123 

Cor.    It  follows  directly  that  under  corresponding  conditions 

the  order  of  differentiation  in  the  higher  partial  derivatives  is 

indifferent. 

dh  dsz        .  ffhs 


E.g., 


dx  dy  dx      dx2  dy      dy  dx2 


EXERCISES 


1.  Verify  that  -^_  =  J£}L.%  when  u  =  xh/\ 

dx  dy      dy  dx 

2.  Verify  that  -_^H_  =  J^,  when  u  =  xhj  +  xy9. 

dx  dy2      dy'2  dx 

3.  Verify  that  -^L  =  -^-,  when  u  =  y  log  (1  +  xy). 

dx  dy      dy  dx 

4.  In  Ex.  3  are  there  any  exceptional  values  of  x,  y  for  which  the 
relation  is  not  true? 

5.  Given  u  =  (x2  +  y2)l,  verify  the  formula 

*2^  + 2*^  +  ^  =  0. 
x2  dx  dy        dy2 

6.  Given  u  =  (xs  -f  t/s)  I,  show  that  the  expression  in  the  left 
member  of  the  differential  equation  in  Ex.  5  is  equal  to  — . 

7.  Given  u  =  (x*  +  y*  +  z*yh  ■  prove  that  B~  +  ^  +  0  =  0. 

8.  Given  u  =  sec  (y+ax)  +  tan  (?/  -  ax)  ;  prove  that  —  =  a^'u- 

dx2         dy2 

9.  Given  M=sinx  cosy;  verify  that     d  u    = ^ =    ^u    . 

J         dy2dx2    dxdydxdy    dx2dy2 

d2u        d2u 

10.  Given  «=(4a&-  c2)-| ;  prove  that  :rr  =  r-^!* 

11.  If  «=*&L,  show  that  o;^+„-i^-=2^. 

*  +  #  dx*     Jdxdy       dx 

12.  Given  u  =  log  (a:2  +  y2),  prove  5^  +  —  =  0. 


a<2  ■  ay 


13.  If  u  =  (xn  +  yny,  show  that  the  equation  of  Ex.  5  is  satisfied. 

14.  Given  u  =  (x*  +  y*  +  z2  +  w2)"1,  prove  ^  +  <*%  +  ^f  +  S^M.  =  0 

d*2     9^      a*2      Qw* 


CHAPTER   IX 

CHANGE    OF   VARIABLE 

69.    Interchange  of  dependent   and  independent  variables.      If 

y   is   a   continuous   function   of   x,   defined   by   the   equation 

f(x,  y)  =  0,  the  symbol  —  represents  the   derivative  of  y  with 
regard  to  x,  when  one  exists.     If  x  is  regarded  as  a  function  of 

y,  defined  by  the  same  equation,  the  symbol  —  represents  the 

dy 

derivative  of  x  with  regard  to  y,  when  one  exists.     It  is  re- 
quired to  find  the  relation  between  -^  and  — 

dx  dy 

Let  x,  y  change  from  the  initial  values  xly  yx  to  the  values 

x1  +  Ax,  yx  +  Ay,  subject  to  the  relation  f(x,  y)  =  0. 

Then,  since  A 

Ay  _  1 

Ax      Ax 

Ay 


it  follows,  by  taking  the  limit,  that 

dx     dx 
dy 


(1) 


Hence,  if  y  and  x  are  connected  by  a  functional  relation,  the 
derivative  of  y  with  regard  to  x  is  the  reciprocal  of  the  derivative 
of  x  with  regard  to  y. 

This  process  is  known  as  changing  the  independent  variable 
from  x  to  y.     The  corresponding  relations  for  the  -higher  de- 

124 


CHANGE   OF   VARIABLE 


125 


rivatives  are  less  simple.     They  are  obtained  in  the  following 

manner : 

d  y  dx   d  x 

To  express  — 4  in  terms  of  — ,  — -   differentiate  (1)  as  to  x ; 
dx2  dy  dy2 


d2y  _  d  [  1 

d 

[  1 

dy  _  d 

\  1  1 

dx2      dx  1  dx 

dy 

dx 

dx      dy 

dx 

idy 

dy\ 

idyl 

dx 
dy 


But 


hence 


In  a  similar  manner, 


—  l1}- 

dy  dx  1 

idyl 

d2x 
dy2  , 
/dx\2 

[dyj 

d?x 

d2y 
dx2 

dy2 

fdx\3' 

\dy) 

ner, 
d3y 

d3x  dx      o  fd2x\i 
d?dy         WJ 

dx3 

fdx\5 

\dy) 

(2) 


(3) 


70.   Change  of  the  dependent  variable.     If  y  is  a  function  of  z, 

let  it  be  required  to  express  —    — ,  •••in  terms  of  — ,  — -,  •••. 

dx   dx-  dx    dx-' 

Suppose  y  =  <f>(z).     Then 

dy  ^dycte  =  ^u\dz_ 
dx      dz  dx  dx 


dx2      dxVKJdx 


dz  d    .i^\,,i/^d2z 


126  DIFFERENTIAL   CALCULUS 


But  ^-<l>'(z)  =  -cp'(z)-  =  <p"(z)  —  ■ 

dz*K  ;      dz     v  ;<to      v  v  "eta 


S-^S'+^S-        (4) 

The  higher  ^derivatives  of  y  can  be  similarly  expressed  in 
terms  of  ^-derivatives  of  z. 

71.    Change  of  the  independent  variable.     Let  y  be  a  function 
of  x,  and  let  both  x  and  y  be  functions  of  a  new  variable  I.     It 

is  required  to  express  —  in  terms  of  -3L.  and  —  in  terms  of 
1      ,  cto  eft'  dx2 

-■-  and  — a- 
eft  eft2 

By  Art.  8,  e% 

%=d*?  (1) 

die      do? 


hence 


d2?/  C?£C      d2^  e% 
^y_~dfdi~~d^di  (9) 


fdx\* 
dt 


In  practical  examples  it  is  usually  better  to  work  by  the 
methods  here  illustrated  than  to  use  the  resulting  formulas. 

72.    Simultaneous  changes  of  dependent  and  of  independent  vari- 
ables.    Suppose,  for  example,  that  an  equation  involving  x,  y, 

— ,  .  .  .  is  given,  and  it  is  required  to  transform  the  equation 

into  polar  coordinates  by  means  of  the  formulas  x  =  p  cos  0, 
y  =  p  sin  6.  Since  the  variables  x  and  y  are  connected  by  some 
equation  (y  being  a  function  of  x),  we  may  regard  x,  y,  p  as 


CHANGE   OF   VARIABLE 

functions  of  6.     E.g.,  consider  the  function 


12' 


11  = 

HtJf 

dry 

dx2 

rom  Art.  71, 

dy 

dy  _d0f 

dx     dx 

dO 

dx     d-y      dy     drx 

d2y      dO  '  d6-      dO  '  d0\ 

dx- 

/dxX3 

tie. 


By  substituting  these  values  in  the  expression  for  E,  it  becomes 


R 


dx\2  ,  (dy 
dO         \d0 


dx     d~y      dy  ^  d2x 
dO  '  ~d&-  ~  dd  '  d¥ 


This  is  in  terms  of  a  new  independent  variable  0.  We  have 
now  to  express  these  ^-derivatives  of  x  and  y  in  terms  of  p 
and  6. 

From  the  relations  x  =  p  cos  0,  y  =  p  sin  0  we  have 

g  =  -pcos*-2sin*|+cos*g, 


dhj 


d$- 


'  =  -  p  sin  0  +  2  cos  0  =c  4-  sin  0 


dp 


d$ 


de2 


128  DIFFERENTIAL   CALCULUS 

Upon  substituting  these  values  in  the  last  expression  for  B, 
we  obtain 


B  = 


W] 


f  +  2(«!>)*-p*E 


(16 


EXERCISES 

1.  Change  the  independent  variable  from  x  to  z  in  the  equation 

x2  — \  +  x  — '-  +  y  =  0,  when  x  =  e*. 
dx2        dx     a 

dx      dz 

dx2      dz2  dz 

Hence  x2^-  +  x^-  +  y  =  0  becomes  ^  +  y  =  0. 

2.  Interchange  the  function  and  the  variable  in  the  equation 

dx2  \dxj 

3.  Interchange  x  and  y  in  the  equation 


</x2 

4.  Change  the  independent  variable  from  ar  to  y  in  the  equation 

JdY\2_dy  d*«_d2y  (djY=  o 
V^2/       dx  dx*     dx2  \dx) 

5.  Change  the  dependent  variable  from  y  to  z  in  the  equation 

!^=  1 +2(1 +.V)fM',  when  y  =  tans. 
</x2  1  +  y*    \dxl  * 


CHANGE   OF    VARIABLE  129 

6.  Change  the  independent  variable  from  x  to  y  in  the  equation 

x2 h  x \-  u  =  0,  when  y  =  log  x. 

dx2         dx 

7.  If  y  is  a  function  of  x,  and  x  a  function  of  the  time  t,  express 
the  ^/-acceleration  in  terms  of  the  ar-acceleration,  and  the  x-velocity. 

Since  dy  =  dydx 

dt      dx  dt 


hence 


d2y  _  dy  (Px  .dx     d  fdy\ 
dt2  ~  dx  dt2       dt      <it\dx)' 

c]l((111\  =^_((J][\flx=d2y(lx 
it\dxl  ~  dxKdxj  dt     dx2  dt' 


But 

dt 


d2y  _  dy  d2x      d2y  ldx\2 

hence  ^-s^+sUi' 


In  the  abbreviated  notation  for  /-derivatives, 

dx  dx2 

8.    Change  the  independent  variable  from  x  to  u  in  the  equation 

dry         2  x     dy  y 

dx2      1  +  x2  dx      (1  +  x2)2 


]y  + y =  0,  when  x  =  tan  u. 


9.    Change  the  independent  variable  from  x  to  t  in  the  equation 

(1  _  X2)  A  _  x  (Il  =  o,  when  x  =  cos  t. 
dx2         d-x 


10.    Show  that  the  equation 

0d2y  ,      dy  .  A 

dx2         dx      J 

remains  unchanged  in  form  by  the  substitution  x  =    . 

EL.    CALC 9 


130  DIFFERENTIAL   CALCULUS 

11.  Interchange  the  variable  and  the  function  in  the  equation 

dx2      \dxl        y\dx) 

12.  Change  the  dependent  variable  from  y  to  z  in  the  equation 

|j*+ a  -  y)%+y2  =  °'  when  y  = z'2- 

Change  the  independent  variable  from  x  to  t  in  the  equations : 

13.  CI  —  x2)  —  —  x—  +  y  =  0,  given  a:  =  cos  t. 
K  clx2         dx      y         '  h 

14.  x3  — -  +  3  x2 \-  x 1-  v  =  0,  given  x  —  el. 

dx3  dx'1         dx 

15.  x4^+fl2^  =  0,  x  =  -. 

dx2  t 

dy 

Xdx~V 

16.  Transform  by  assuming  x  =  p  cos  6,  y  =  p  sin  0. 

17.  Given  z  =  7  +  *2,  ?/  =  3  +  *2  -  3  l\     Find  — . 


CHAPTER  X 

EXPANSION  OF  FUNCTIONS 

It  is  sometimes  necessary  to  expand  a  given  function  in  a 
series  of  powers  of  the  independent  variable.  For  instance, 
in  order  to  compute  and  tabulate  the  successive  numerical 
values  of  sin  x  for  different  values  of  x,  it  is  convenient  to 
have  sin  x  developed  in  a  series  of  powers  of  x  with  coeffi- 
cients independent  of  x. 

Simple  cases  of  such  development  have  been  met  with  in 
algebra.     For  example,  by  the  binomial  theorem, 

(a  +  x)n  =  an  +  nan~lx  +  n0l -1)  a— v  +  • . . ;  (1) 

J.  •  A 

and  again,  by  ordinary  division, 

j-L-=l  +  s  +  3»  +  aj»+....  (2) 

It  is  to  be  observed,  however,  that  the  series  is  a  proper 
representative  of  the  function  only  for  values  of  x  within  a 
certain  interval.  For  instance,  the  identity  in  (1)  holds 
only  for  values  of  x  between  —  a  and  -f  a  when  n  is  not  a 
positive  integer ;  and  the  identity  in  (2)  holds  only  for  values 
of  x  between  —  1  and  -f  1.  In  each  of  these  examples,  if  a 
finite  value  outside  of  the  stated  limits  is  given  to  x,  the  sum 
of  an  infinite  number  of  terms  of  the  series  will  be  infinite, 
while  the  function  in  the  first  member  will  be  finite. 

131 


132  DIFFERENTIAL   CALCULUS 

73.  Convergence  and  divergence  of  series.*  An  infinite  series 
is  said  to  be  convergent  or  divergent  according  as  the  sum  of  the 
first  n  terms  of  the  series  does  or  does  not  approach  a  finite 
limit  when  n  is  increased  without  limit. 

Those  values  of  x  for  which  a  series  of  powers  of  x  is  con- 
vergent constitute  the  interval  of  convergence  of  the  series. 

For  example,  the  sum  of  the  first  n  terms  of  the  geometric 

series  ,         ,       2  ,       s  , 

a  -f-  ax  4-  axz  +  ax?  -\ 

ail  —  xn) 
is  sn  =  -1 >-. 

1  —  x 

First  let  x  be  numerically  less  than  unity.  Then  when  n  is 
taken  sufficiently  large,  the  term  xn  approaches  zero ; 

hence  Hm  sn  =  -^-- 

W  =  co  l_x 

Next  let  x  be  numerically  greater  than  unity.      Then  xn  be- 
comes infinite  when  n  is  infinite ;  hence,  in  this  case 

lim 
,,-od  *»  =  «>• 

Thus  the  given  series  is  convergent  or  divergent  according 
as  x  is  numerically  less  or  greater  than  unity.  The  condition 
for  convergence  may  then  be  written  . 

-l<a<i, 

and  the  interval  of  convergence  is  between  —  1  and  -|-  1. 
Similarly  the  geometric  series 

l-3z  +  9z2-27^+...,> 

*  For  an  elementary,  yet  comprehensive  and  rigorous,  treatment  of  this 
subject,  see  Professor  Osgood's  "  Introduction  to  Infinite  Series"  (Harvard 
University  Press,  1897). 


EXPANSION  OF  FUNCTIONS  133 

whose  common  ratio  is  —  3  x,  is  convergent  or  divergent  accord- 
ing as  3  a;  is  numerically  less  or  greater  than  unity. 

The  condition  for  convergence  is  —  1  <  3  x  <  1,  and  hence 
the  interval  of  convergence  is  between  —  ^  and  -f-  ^. 

74.   General  test  for  convergence. 

Let  £  =  ?<!  +  u2  +  u3-\ hMn  +  MB+1  H 

be  a  series  of  positive  terms  having  the  property  that    w+1  <  r 

(r  a  fixed  proper  fraction)  for  all  values  of  n  that  exceed  a  def- 
inite integer  k  that  can  be  assigned.  We  wish  to  prove  that 
under  these  conditions  S  is  convergent.  This  is  called  the  ratio 
test  for  convergence. 

According  to  hypothesis  we  have  the  inequalities 

^±l<r,   ^*±*<r,   ^±-3<r,  etc. 

Uk  Uk+\  Uk+2 

By  multiplying   the  first  two  equalities  together  we  obtain 

^^  <  r2;    then,  multiplying  this  result  by  the  third  of  the 
uk 

given  inequalities  we  deduce  further  -^±?<r3;  and  so  on.    These 

results  may  be  written  in  the  form 

%+i  <  ruk,  uk+2  <  r*u k,  w4+3  <  rhik,  .-• ,  uk+p  <  r'uk. 

Hence  we  have  the  inequality 

S  <  ux  -f  u2  H (-  uk  -f  ruk  -f  r2^  +  rX  -\ 

But  the  series  in  the  right  member,  which  may  be  denoted  by 
S'j  can  be  put  in  the  form 

S'  =  ^  +  W2 -f  —  +  m*_i  +  uh  (1  -f  r  +  r2  +  r3  +  •••) 

=  wx  +  w2  H f-  ?^_!  +  -^*-  • 

1— r 


134  DIFFERENTIAL   CALCULUS 

The  terms  ux,  u2,  ••• ,  uk  being  assumed  finite,  it  follows  that 
S'  is  finite  and  hence  S,  which  is  less  than  S',  also  is  finite. 
Since  S  is  formed  by  the  successive  addition  of  positive  terms, 
it  follows  that  the  series  S  converges  towards  a  definite  finite 
limit. 

If  the  series  S  contains  an  infinite  number  of  negative,  as 
well  as  of  positive,  terms,  it  converges  whenever  the  series 
formed  by  the  positive,  or  absolute,  values  of  its  terms  eon- 
verges.     The  series  is  then  said  to  be  absolutely  convergent. 

In  order  to  prove  the  preceding  theorem,  we  obeerve  that 
the  positive  terms  of  S  taken  alone  form  a  converging  series, 
whose  limit  will  be  denoted  by  P,  and  the  negative  terms  taken 
alone  will  form  a  converging  series  whose  limit  will  be  denoted 
by  —  J¥.t  Let  Sm  denote  the  sum  of  the  first  m  terms  of  S  and 
suppose  that  these  consist  of  p  positive  terms  whose  sum  is 
denoted  by  Pp  and  of  n  negative  terms  whose  sum  is  —  Nn. 
Then  we  have  Sm=  Pp  —  Nn.  Now  when  m  becomes  infinite, 
p  and  n  also  become  infinite,  and  hence 

a  lim     -  e    .  .     Km       p  _    lira       jy  __  p  _  *r 

m  i  oo      m      p  =  oo       p      n  =  oo       » 

Therefore,  S  is  convergent. 

When  a  series  is  convergent,  but  the  series  formed  with 
the  absolute  values  of  its  terms  is  not  convergent,  the  given 
series  is  said  to  be  conditionally  convergent* 

The  absolute  value  of  a  real  number  u  is  its  numerical  value 
taken  positively,  and  is  written  |  u  |. 

If  a  series  consists  of  terms  that  are  alternately  positive 

*  The  appropriateness  of  this  terminology  is  due  to  the  fact  that  the  terms 
of  an  absolutely  convergent  series  can  be  rearranged  in  any  way, "without 
altering  the  limit  of  the  sum  of  the  series  ;  and  that  this  is  not  true  of  a  con- 
ditionally convergent  series.    For  a  simple  proof,  see  Osgood,  pp.  43,  44. 


EXPANSION  OF  FUNCTIONS  135 

and  negative,  and  if,  after  any  definite  term  of  the  series,  each, 
succeeding  term  is  numerically  less  than  the  preceding  one, 
then  the  series  is  convergent. 

For,  suppose  that  beginning  with  the  term  nk,  the  series  is 

&'  =  Vk  ~   Vk  +  l  +  %+2  —  Uk+3  +  Uk+i  ~   '"  > 

in  which  uk,  vk+l,  etc.  represent  positive  numbers  and  nk+l  <  uk, 
uk+2  <  Uk+u  ••'  j  nm+i  <  l'mi  f°r  every  value  of  m  greater  than  7c. 
By  grouping  the  terms  in  pairs,  (uk  —  iik+l),  (uk+2—uk+^),  ••• ,  each 
of  which  is  positive,  it  is  seen  that  S'  has  a  positive  value, 
which  .may  be  finite  or  infinite. 

But  S'  may  also  be  written  in  the  form 

&  =  uk  -  [(%+i  -  *<*+2)  +  (m*+8  -  uk+i)  +  •••], 

wherein  the  terms  in  brackets  are  all  positive,  hence  S'  has  a 
value  less  than  uk.  It  therefore  converges  towards  a  definite 
finite  limit. 

It  now  follows  that  the  approximate  value  of  S'  obtained  by 
algebraically  adding  uk,  uk+l,  ••• ,  um  differs  from  the  true  value 
of  the  series  by  a  number  less  than  um.  This  fact  can  be 
shown  in  precisely  the  same  way  as  that  by  which  S'  has  just 
been  shown  to  have  a  value  less  than  uk. 

Ex.  1.     Is    the    series    1 1 \-  ...  +  (—  l)'1-1  -  +  •••  con- 

2      3      4  n 

vergent  i 

Since  the  terms  are   alternately  positive  and  negative  and  their 

numerical  values  are  always  decreasing,  it  follows  at  once  from  the 

preceding  paragraph  that  this  series  is  convergent.     It  will  be  found 

later  that  its  value  is  log  2. 

Ex.  2.     Prove  the  convergence  of  the  series  met  with  in  Art.  16, 

a+i.+l  +  i+...+J-+...-. 

•2!      31      4!T  n\ 


136  DIFFERENTIAL   CALCULUS 

In  this  case  un  =  1 ,  wn+1  = 1— .     Hence  ^±1  =  — 1—  .    This 

n  !  (w  +  1) !  m„        n  +  1 

ratio  is  less  than  \  for  all  values  of  n  greater  than  2,  and  the   ratio 

condition  for  convergence  is  satisfied. 

Ex.  3.     Prove  the  divergence  of  the  harmonic  series 

The  ratio  un+i :  un  becomes  greater  than  r  when  n  is  sufficiently 
large.     By  grouping  the  terms  it  may  be  written  in  the  form 

1 +  *+(*  +  *)  +  (*  + *  +  *  +  »+-, 
the  succeeding  groups  having  23,  24,  ••• ,  2n,  •••  consecutive  terms  re- 
spectively.    The  sum  of  the  terms  in  any  group  is  greater  than  \. 
For,  in  the  nth  group  the  last  term  —  has  the  least  value,  and  as  there 

are  2n_1  terms  in  the  group  their  sum  is  greater  than  2n_1    —  =  -• 
As  there  is  an  infinity  of  such  groups,  their  sum  is  infinite. 

Ex.  4.     The  series 

£=1  +  1  +  1-+  ...  +1+... 

is  convergent  for  p  >  1. 

Let  the  terms  of  S  be  grouped  in  the  following  manner : 

the   nth   group  beginning  with  and  containing  2n_1  terms. 

The  nth  group  is  accordingly  less  than  its  first  term  multiplied  by 

the  number  of  terms  in  the  group,  that  is,  <  2n_1  •  —  =  — — . 

&       r'  '  (2"_1)p     (2n— 1)p_1 

Hence  we  deduce  the  inequality 

the  right  member  of  which  is  a  geometric  series  having  — —  as  the 
common  ratio.     It  is  therefore  convergent,  and  hence  S  is  convergent, 


EXPANSION   OF   FUNCTIONS  137 

if    <  1.     This  inequality  is  satisfied  for  every  value  of  ^greater 

than  unity.     Moreover,  it  was  shown  in  Ex.  3  that  ior  p  =  1  the  series 
5   is    divergent.      When  p  <  1,  S   is    divergent.     For   in   that   case 

— ->-,  n    is   any    positive    integer    (except    1),   and    therefore    the 
np      n 

terms  of  S  are  greater  than  the  corresponding  terms  of  the  harmonic 

series. 

Hence : 

The  necessary  and  sufficient  condition  that  the  series  1  -\ —  +  - \-  ••• 

may  converge  is  p  >  1. 

Ex.  5.    Show  that  the- series 1 = — \- 


12      2-3      3.4+  n(n+l) 

is  convergent. 

This  may  fc>e  proved  by  comparison  with  the  series  in  Ex.  4  for  the 

particular  case  p  =  2. 

1      .-,       1      .1        1      .1  1  .1 


Since 


l-2<    '  2-3<22'  3-4<32'  '"  '  n(n  +  1)  <n*'  "' 


it  follows  that  the  value  of  the  given  series  is  less  than  that  of 

which  is  known  to  be  convergent  on  account  of  the  theorem  deduced  in 
the  preceding  example. 

Ex.  6.   Examine  for  convergence  the  series  whose  nth  term  is  — — ■ 

n2+l 

r  n  =  *  ^  -1 

Hint.  n2  + 1  1  ^  n  +  1 

n  -\ 
n 


Ex.  7.   Examine  for  convergence  the  series 

l_2fi  _  .      ,   (~l)w~^ 
2      5      10  n2  +  1 


Ex.  8.     Determine  whether  the  series  whose  nth  term  is 


n2  +  1 

convergent  or  not;  the  series  whose  general  term  is  — —  . 

n3  +  l 


138  DIFFERENTIAL   CALCULUS 

75.  Interval  of  convergence.  If  the  terms  u1}  u2,  •••  of  a  given 
series  are  functions  of  a  variable  x,  then  the  series  will  usually 
converge  for  .some  values  of  x  and  diverge  for  all  others.  In 
such  a  case  the  problem  is  to  determine  the  interval  of  conver- 
gence, that  is,  the  range  of  values  of  x  for  which  the  series  is 
convergent  The  following  examples  will  illustrate  the  method 
of  procedure. 

Ex.  1.     Determine  the  interval  of  convergence  of  the  series 
1  +  x  -f  2  x2  +  3  xs  +  •••  +  nxn  +  ...  . 
In  this  case  un  =  (n  —  l)^""-1  and  un+1  =  nxn.. 

Hence,  ^  =         "*" =  -JL-  x. 

un         (n-l)xn-1      n-1 

According  to  the  ratio  condition  for  convergence,  it  is  necessary 
that  this  ratio  shall  be  numerically  less  than  1  for  all  values  of  n 
exceeding   a  fixed    number    k.      As  n  increases,   the  fraction  

71—1 

approaches  unity.  Hence  if  [  x  \  has  any  fixed  value  less  than  1,  the 
given  series  is  absolutely  convergent.  The  interval  of  convergence 
is  defined  by  the  inequalities  —  1  <  x  <  1. 

It  is  evident  from  the  preceding  example  that  the  ratio  con- 
dition for  the  absolute  convergence  of  a  series  ma}^  be  written 


lim 


l±n±l 


<  1,  (3) 

which  is  especially  convenient  for  application. 

Ex.  2.     Find  the  interval  of  convergence  of  the  series 
1  +  2  •  2  x  +  3  •  4  x2  +  4  •  8  x*  +  5  •  16  x*  +.-  . 


EXPANSION   OF   FUNCTIONS  139 

Here  the  nth  term  un  is  n  2n_1xn_1,  and  the  (n  -f  l)th  term  un+1  is 

(n  +  \)2nxn; 

i                                 ^n+i      (n  4-  l)2"z*      (n  + 1)  2  x 
lience  -J!±±=*-— — ^ =  i — — i 

wn         w2n-1x"  »  n 

therefore  when  n  =  co,  -^  =  2  x. 

It  follows  by  (3)  that  the  series  is  absolutely  convergent  when 
—  1  <2#<1,  and  that  the  interval  of  convergence  is  between  —  \ 
and  +  \.  The  series  is  evidently  not  convergent  when  x  has  either 
of  the  extreme  values. 

Ex.  3.     Find  the  interval  of  convergence  of  the  series 


x 
1-3      3 


x3  x5  x"  >      (_l)n-i^2M-i      ^ 

•  33      5-35     7-37      ""      (2n-l)32"-1 


Here 


t/H4.i  I  _  2n  -  1     32""*     a;2"*1  =  2n-l     z2. 
wft  2  n  +  1  '  b2"*1  '  x2"-i      2  n  +  1  '  32' 


v  Inn 

hence 

n  =  co 


32' 

a:2 
thus   the   series   is   absolutely   convergent   when   —  <  1,   i.e.,   when 

b 

—  3  <  x  <  3,  and  the  interval  of  convergence  is  from  —  3  to  +  3. 
The  extreme  values  of  x,  in  the  present  case,  render  the  series  con- 
ditionally convergent. 

Ex.  4.     Determine  the  interval  of  convergence  of  the  series 


2 !      4 !      6 !  .  '       (2  n  -  2) ! 

Since  even  powers  of  x  are  positive,  the  terms  of  this  series  are 
alternately  positive  and  negative.     The  term  wn+1  is  derived  from  un 

by  multiplying  it  by .     For  all  values   of   n    such    that 

(2  n  —  1)2  n 

this  fraction  is  less  than  1,  we  shall  have  the  condition  [  un+1  |  <  |  nn  \ 

and  the  series  is  convergent  on  account  of  the  property  of  series  with 

alternately  positive  and  negative  terms. 


140  DIFFERENTIAL   CALCULUS 

Ex.  5.     Prove  the  convergence  of  the  series 

Inthiscase  |  un  |=z2-  f1  +-+•••  +  — Y     Notice  that    h.m    I  u  J 
\9      4  2"-1/  n=<x>  '    n[ 

is  not  zero.     The  series  is  nevertheless  convergent,  but  not  absolutely 

convergent. 

Ex.  6.     Determine  the  interval  of  convergence  for  the  series 

16  n 

Ex.  7.     Determine  the  interval  of  convergence  for  the  series 

-A_+        2  3        +...+_n      _+.... 

x-1      (x-1)2      (ar-l)*  (ar-l)w 

Ex.  8.     Find  the  interval  of  convergence  for  the  binomial  series 

-,    ,         ,   a(a  —  1)    9  ,  a(a  —  Y)(a  —  2)    „  , 

1  4-rtxH — i — *-  x2  H — ^ ^ '-  x3  + 

in  which  a  is  any  constant. 
Ex.  9.     Show  that  the  series 

has  the  same  interval  of  convergence  as  that  of  Ex.  3 ;  but  that  the 
extreme  values  of  x  render  the  series  absolutely  convergent. 

76.  Remainder  after  n  terms.  The  last  article  treated  of 
the  interval  of  convergence  of  a  given  series  without  reference 
to  the  question  whether  or  not  it  was  the  development  of  any 
known  function.  On  the- other  hand,  the  series  that  present 
themselves  in  this  chapter  are  the  developments  of  given  func- 
tions, and  the  first  question  that  arises  is  concerning  those  values 
of  x  for  which  the  function  is  equivalent  to  its  development. 


EXPANSION   OF   FUNCTIONS  141 

When  a  series  1ms  such  a  generating  function,  the  difference 
between  the  value  of  the  function  and  the  sum  of  the  first  n  terms 
of  its  development  is  called  the  remainder  after  n  terms.  Accord- 
ingly, if  f(x)  is  the  function,  Sn(x)  the  sum  of  the  first  n  terms 
of  the  series,  and  lin(x)  the  remainder  obtained  by  subtracting 
S„(x)  from  fix),  then 

in  which  Sn(x),  Rn(x)  are  functions  of  n  as  well  as  of  x. 

If  JToO  R-^)  =  0>     tllei1     Hi"      «-(*)  =/(*)  i 

thus  the  limit  of  the  series  Sn(x)  is  the  generating  function 
when  the  limit  of  the  remainder  is  zero.  Frequently  this  is 
a  sufficient  test  for  the  convergence  of  a  series. 

If  a  series  is  expressed  in  integral  powers  of  x  —  a,  the  pre- 
ceding conditions  are  to  be  modified  by  substituting  x  —  a  for 
x ;  in  other  respects  each  criterion  is  to  be  applied  as  before. 

77.    Maclaurin's   expansion  of  a  function  in  a  power-series.* 

It  will  now  be  shown  that  all  the  developments  of  functions 
in  power-series  given  in  algebra  and  trigonometry  are  but 
special  cases  of  one  general  formula  of  expansion. 

It  is  proposed  to  find  a  formula  for  the  expansion,  in 
ascending  positive  integral  powers  of  x  —  a,  of  any  assigned 
function  which,  with  its  successive  derivatives,  is  continuous 
in  the  vicinity  of  the  value  x  =  a. 

The  preliminary  investigation  will  proceed  on  the  hypothe- 
sis that  the  assigned  function  f(x)  has  such  a  development, 

♦Named  after  Colin  Maclaurin  (1608-1746),  who  published  it  in  his 
"  Treatise  on  Fluxions  "  (1742) ;  but  he  distinctly  says  it  was  known  by 
James  Stirling  (1692-1770),  who  also  published  it  in  his  "  Methodus  Differ- 
entialis  "(1730),  and  by  Taylor  (see  Art.  78). 


142  DIFFERENTIAL   CALCULUS 

and  that  the  latter  can  be  treated  as  identical  with  the  former 
for  all  values  of  x  within  a  certain  interval  of  equivalence  that 
includes  the  value  x  =  a.  From  this  hypothesis  the  coeffi- 
cients of  the  different  powers  of  x  —  a  will  be  determined.  It 
will  then  remain  to  test  the  validity  of  the  result  by  finding 
the  conditions  that  must  be  fulfilled,  in  order  that  the  series 
so  obtained  may  be  a  proper  representation  of  the  generating 
function. 

Let  the  assumed  identity  be 

fix)  =  A+B(x-a)  +  C(x-  a)2  +  D(x-  a)3 

+  E(x-ay+-,  (1) 

in  which  A,  B,  C,  •••  are  undetermined  coefficients,  indepen- 
dent of  x. 

Successive  differentiation  with  regard  to  x  supplies  the 
following  additional  identities,  on  the  hypothesis  that  the 
derivative  of  each  series  can  be  obtained  by  differentiating  it 
term  by  term,  and  that  it  has  some  interval  of  equivalence 
with  its  corresponding  function  : 

f(x)  =  B  +  2C(x-a)+      3D(x-a)2+  ±E(x-af+  ••• 

f"(x)=         2(7  +3-2D(x-a)  +      4 .  3E(x-af+  •-. 

f"(x)=  3.2Z>  +4.3.2^(»-o)+-.- 

If,  now,  the  special  value  a  is  given  to  x,  the  following 
equations  will  be  obtained: 

f(a)=A,f(a)  =  B,  /"(a)  =  2  C,f"(a)  =  3-2D,  -. 

Hence, 

A  =/(«),  B=/(a),  C=f-^f,  D=f-^,  ■- 


EXPANSION   OF   FUNCTIONS 


143 


The  coefficients  in  (1)  are  now  determined,  and  the  required 
development  is 

/(a5)=/(a)+/(a)(a5-a)+^^(aj-a)«+£^(a5-a)» 

(2) 


2! 


This  is  known  as  Maclaurin's  series,  and  the  theorem  ex- 
pressed in  the  formula  is  called  Maclaurin's  theorem. 


Ex.  1.     Expand  log  x  in  powers  of  x  —  a,  a  being  positive 

1 
x 


Here      /(*)  =  log  x,f(x)  =  £,/"(*) 


1  1.0 

V'"(*)=:LT 


Hence,        /(a)  =  log  a,f'(a)  =  If"  (a)  =  -  I,  /"'(a)  =  ^ 


1-2 


/«(a) 

and,  by  (2),  the  required  development  is 


(-l)-i(w-l)| 


log  x  =  log  a  +  1  (*  -  a)  -  J-  O  -  a)2  +  -L  (a;  -  a)J 
a  2  a2  o  a3 


The  condition  for  the  convergence  of  this  series  is 


lira   I  (x  —  a)^1   .  (x  -  a)> 


(n  +  l)an+1 

x  —  a 


nan 


<1, 


<1, 


\x  -  a  \<a, 

0<x<2a. 

•     This  series  may  be  called  the  development  of  log  x  in  the  vicinity  of 
x  —  a.     Its  development  in  the  vicinity  of  x  =  1  has  the  simpler  form 

log  X  =  X  -  1  -  1(X  -  I)*  +  \(X  -  l)3  -   .., 

which  holds  for  values  of  x  between  0  and  2. 


144  DIFFERENTIAL   CALCULUS 

In  using  this  series  for  the  computation  of  a  table  of  logarithms 
we  may  put  for  a  any  number  whose  logarithm  is  already  known,  and 
for  x  any  number  near  a  in  magnitude.  It  is  a  great  advantage  to 
keep  x  —  a  so  small  that  the  power-series  in  x  —  a  may  be  not  merely 
convergent,  but  may  converge  to  its  limit  so  rapidly  that  all  powers  of 
x  —  a  above  the  fourth  or  fifth  may  be  neglected  without  affecting  the 
desired  degree  of  accuracy. 

E.g.,  being  given  log  10  =  2.302585,  suppose  it  is  required  to  com- 
pute log  11,  log  12,  •••,  log  20.     Put  a  =  10,  and  x  =  11.     Then 

logi^iogio+^-Kr^HKi^^K^+Ki^-Ki^HKiV)7--- 

The  numerical  work  may  be  tabulated  in  the  following  form : 

+  2.30258509 

.10000000  -  .00500000 

.00033333  .00002500 

.00000200  .00000017 


.00000001  -  .00502517 


2.40292043 
-  .00502517 


2.39789526 
Hence  log  11  =  2.397895-., 

correct  to  six  places  of  decimals.  To  make  sure  of  the  sixth  figure  it 
is  well  to  carry  the  work  to  seven  or  eight  figures.  The  remaining 
terms  of  the  series  after  KrV)7  cannot  affect  this  result,  because  their 
sum  is  less  than  an  infinite  decreasing  geometric  progression  whose 
first  term  is  IGV)8  and  whose  ratio  is  fo.     From  the  formula 

1  -  r 

i 
it  follows  that  the  remainder  is  less  than 


72  •  109 

To  calculate  log  12,  log  13,  •••  we  could  now  keep  a  =  10  and  let 
x  =  12,  13,  •••  successively,  but  in  order  to  secure  rapid  convergence  it 
is  better  to  change  the  value  of  a,  choosing  for  a  the  nearest  number 


EXPANSION  OF  FUNCTIONS  145 

whose  logarithm  has  been  found.  Thus,  in  computing  log  12  we  can 
use  either  of  the  two  series 

log  12  =  log  io  +  a  -  KA)2+  K&Y  -  -. 
log  12  =  log  li  +  A  -  KA)2  +  Kt1!)3  -  - ; 

but  it  will  be  found  that  five  terms  of  the  second  furnish  as  close  an 
approximation  as  nine  terms  of  the  first.  The  practical  advantage 
of  the  step-by-step  process  will  depend  on  how  many  of  the  intermedi- 
ate values  we  actually  require.  If  we  are  given  log  10  and  wish  to 
compute  log  15,  it  may  be  easier  to  compute  the  latter  directly  with- 
out determining  the  intermediate  values. 

Ex.  2.  Develop  f(x)  =  xs  —  2  x1  +  5  x  —  7  in  powers  of  x  —  1  and 
use  the  result  to  compute/(1.02),/(1.01),/(.99),/(.98). 

Ex.  3.     Develop  3  y2  —  14  y  +  7  in  powers  of  y  —  2. 

Ex.  4.  Expand  sin  x  in  powers  of  x  —  a  and  use  the  result  to  com- 
pute sin  31°. 

Let  a  =  30°,  x  =  31°.     In  the  formula 

sin  x  =  sin  a  +  cos  a(x  -  a)-^^(x  -  a)2  -  ^^  (x  -  a)3  •», 

the  difference  x  —  a  becomes  1°  or  .001745  radians,  and  the  coefficients 
of  its  various  powers  are  all  known ;  since  sin  a  =  .5,  cos  a  =  .866025  the 
work  is  now  reduced  to  numerical  calculation  in  which  three  terms 
are  sufficient  to  obtain  the  result  correct  to  six  places  of  decimals.  In 
general,  to  calculate  sin  x  or  cos  x,  take  for  a  the  nearest  value  for 
which  sin  a,  cos  a  are  known. 

The  expansion  of  a  function  f(x)  in  a  series  of  ascending 
powers  of  x  can  be  obtained  at  once  from  formula  (2)  by  giv- 
ing a  the  particular  value  zero.     The  series  then  becomes 

A*)=A0)+/-(0)x+q|i^+...+^^+....    (3) 

EL.    CALC.  —  10 


146                              DIFFERENTIAL  CALCULUS 

Ex.  5.   Expand  sin  x  in  powers  of  x,  and  find  the  interval  of  con- 
vergence of  the  series. 

Efeve            /0)  =  sin  x,  /(0)  =  0, 

f(x)  =cosx,  /'(())  =1, 

f"(x)  =  -  sin  a;,  /"(0)  =  0, 

/'"(*)  =  - cos*,  /"'(())=  -1 

f"(x)  =  sinx,  /iv(0)=0, 

/v(ar)  =  cos  x,  /v(0)  =  1, 


Hence,  by  (3), 


0  +  1  •  x  +  0  •  x2  -  i.  x-3  +  0  •  x4  + 
o ! 


thus  the  required  development  is 

sin  a:  =  a; xs  H x5 x7  +  •••  + -^ — 2!! — a:2"-1  +  .... 

3!  5!         7!  (2w-l)! 

To  find  the  interval  of  convergence  of  the  series,  use  the  method 
of  Art.  74.     The  ratio  of  un+1  to  un  is 

un+\_       a>+1  Z2""1  x2 


«n       (2n  +  l)!     (2n-l)l       (2n  +  l)2n 

This  ratio  approaches  the  limit  zero,  when  n  becomes  infinite,  how- 
ever large  may  be  the  fixed  value  assigned  to  x.  This  limit  being  less 
than  unity,  the  series  is  convergent  for  any  finite  value  of  x,  and 
hence  the  interval  of  convergence  is  from   —  oo  to   -f  oo. 

The  preceding  series  may  be  used  to  compute  the  numerical  value 
of  sin  x  for  any  given  value  of  x.  It  is  rapidly  convergent  when  x  is 
small.     Take,  for  example,  x  =  .5  radians.     Then 

sin(.5)_.o-— +2>3>4>5      2.3.^.5.6.7+  -> 


EXPANSION   OF   FUNCTIONS  147 

=  .5000000 

-  .0208333 
+  .0002604 

-  .0000015 
+  .0000000 


sin  (.5)  =  .4791256  •• 


Show  that  the  ratio  of  u5  to  uA  is  2\-g ;  and  hence  that  the  error  in 
stopping  at  u4  is  numerically  less  than  ut  [^i?  +  Ciis)^  ••*]»  ^na^ 
is,  <  ?fa  u4. 

When  x  is  not  small,  it  is  better  to  use  the  more  general  series  in 
powers  of  x  —  a. 

Ex.  6.     Show  that  the  development  of  cos  x  is 

cosx  =  l  -JL  +  ±--±-+  ...  +i ii ± +  .... 

2!      4!      6!  (2n-2)!  ' 

and  that  the  interval  of  convergence  is  from  —  oo  to  +  oo. 

Ex.  7.     Develop  the  exponential  functions  ax,  ex. 

Here 
/(*)  =  «*,  /'(*)=a*loga,  /"(*)  =  a*(log  «)2,  .»,  /*■>(*)  =a-(loga)»; 
hence       /(0)=l;  /'(0)=log  a,/"(0)  =  (loga)2,  ...,/W(0)  =  (log  a)», 

and  a-H  (log  a)*  +  ^a)V  +  ...  +  (logq)V  +  .... 

•2!  n ! 

As  a  special  case,  put  a  =  e. 
Then  log  a  =  log  e  =  1, 

and  ex^i  +  ^  +  rL  +  rL-f  ...  +£!+  .... 

2.!      3!  n! 

These  series  are  convergent  for  every  finite  value  of  x. 

Ex.  8.     Compute  10*  when  x  =  .1. 
Ex.  9.     Compute  10*  when  r  =  2.01. 


148  DIFFERENTIAL   CALCULUS 

Ex.  10.  Defining  the  hyperbolic  cosine  and  the  hyperbolic  sine 
by  the  equations 

cosh  x  =  \(ex  +  e  x),   sinh  x  =  \{ex  —  e~x), 

prove  —  cosh  x  =  sinh  x,    —  sinh  x  =  cosh  x, 

dx  dx 

cosh  0=1,  sinh  0=0;  and  hence  expand  cosh  x  and  sinh  x  in  powers 
of  x.  Verify  that  cosh  x  +  sinh  x  =  ex,  and  cosh  x  —  sinh  x  =  e~x. 
Compute  cosh  2  and  sinh  2  to  four  decimal  places.  Show  that  the 
error  made  in  stopping  the  series  at  any  term  is  much  less  than  the 
last  term  used. 

78.  Taylor's  series.  If  a  function  of  the  sum  of  two  num- 
bers a  and  x  is  given,  f(a  +  x),  it  is  frequently  desir- 
able to  expand  the  function  in  powers  of  one  of  them, 
say  x. 

In  the  function  /(a  +  x),  a  is  to  be  regarded  as  constant,  so 
that,  considered  as  a  function  of  x,  it  may  be  expanded  by 
formula  (3)  of  the  preceding  article.  In  that  formula,  the 
constant  term  in  the  expansion  is  the  value  which  the  func- 
tion has  when  x  is  made  equal  to  zero,  hence  the  first  term 
in  the  expansion  of  f(a  4-  x)  may  be  written  /(a).  In  the 
same  manner  the  coefficients  of  the  successive  powers  of  x 
are  the  corresponding  derivatives  of  f(a  +  x)  as  to  x,  in  which 
x  is  put  equal  to  zero  after  the  differentiation  has  been  per- 
formed.    The  expansion  may  therefore  be  written 


f(a  +  x)  =  f(a)  +  f(a)x  +£^r-x2  +  •••  +f<    [a)  ocn  +  •••• 

9.  '  m  T 


This  series,  from  the  name  of  its  discoverer,  is  known  as 
Taylor }s  series,  and  the  theorem  expressed  by  the  formula  is 
known  as  Taylor's  theorem. 


EXPANSION  OF  FUNCTIONS  149 


Ex.  1.  Expand  sin  (a  +  x)  in  powers  of  x. 

Here  f(a  +  x)  —  sin  (a  +  x), 

hence  f(a)  =  sin  a, 

and  f'(a)  —  cos  a, 


Hence  sin  (a  +  x)  =  sin  a  +  cos  a  •  x  -  si^x2  _  cos_a   3 

v  '  2!  3! 

Ex.  2.     Compute  sin  61°,  by  putting  a  =  60°. 


EXERCISES 

1.  Expand  tan  x  in  powers  of  x.     Obtain  three  terms. 

2.  Compare  the  expansion  of  tan  x  with  the  quotient  derived  by 
dividing  the  series  for  sin  x  by  that  for  cos  x. 

3.  Find  a  limit  for  the  error  which  occurs  in  replacing  cos  x  by 
the  first  three  terms  of  its  expansion  in  powers  of  x  when  x  =  \  of  a 
radian. 


4.  Prove  that  log  (x  +  Vl  +  x2)  =  x  -  -±—  +  -^-   • 

2-32-4 

5.  Provelogc'os^-^-^-1-^6-?!^!.... 

8  2        4!  6!  8! 

6.  Compute  sin  1°  cdrrect  to  six  places  of  decimals. 


7.  Expand  Vl  —  x2  in  powers  of  x,  and  compare  with  the  expan- 
sion by  the  binomial  theorem. 

8.  Expand  cos  x  in  powers  of  x  —  -. 

9.  Expand  ex+h  in  powers  of  h. 

10.  Arrange  3  x3  —  5  x2  +  8  x  —  5  in  powers  of  x  —  2. 

11.  Expand  log  (x  +  K)  in  powers  of  h. 

12.  Arrange  x4  —  1  in  powers  of  x  +  1. 

X"~N  w  !    Cln~rXr 

13.  Prove  the  binomial  theorem  (a  +  x)n  =  an  +  •••  =  x,- 7~, \T- 

r=0        v  J 

Find  the  form  of  the  series  when  n  is  not  an  integer,  and  determine 
the  interval  of  convergence. 


150  DIFFERENTIAL   CALCULUS 


14.  Find  V12G  =  V125  +  1  =  5  vl  +  T^3  to  three  places  of  deci- 
mals by  the  binomial  theorem. 

15.  Find  S^Lm 

16.  Calculate  log  31. 

17.  What  is  the  greatest  value  of  m  that  will  permit  the  approxi- 
mation (1  +  m)4  =  1  +  4  m  with  an  error  not  exceeding  .001  ? 

18.  Expand  -  in  powers  of   x  —  1  and  find  the  interval  of  con- 

x 
vergence. 

79.  Rolle's  theorem.  By  Art.  76  a  series  can  be  the  correct 
representation  of  its  generating  function  only  when  the  re- 
mainder after  n  terms  can  be  made  as  small  as  desired  by 
taking  n  large  enough.  Before  obtaining  the  form  of  this 
remainder  it  is  necessary  to  introduce  the  following  lemma. 

Rolle's  theorem.  If  f(x)  and  its  first  derivative  are  continu- 
ous for  all  values  of  x  between  a  and  b,  and  if  f(a),  f(b)  both 
vanish,  then  f'(x)  will  vanish  for  some  value  of  x  between  a 
and  b.  * 

The  proof  follows  immediately  from  the  theorem  of  mean 
value  (Art.  39).     See  Figure  41. 

80.  Form  of  remainder  in  Maclaurin's  series.  Let  the  re- 
mainder after  n  terms  be  de- 
noted by  Bn(x,  a),  which  is 
a  function  of  x  and  a  as  well 
as  of  n.  Since  each  of  the 
succeeding  terms  is  divisible 
by  (x  —  a)n,  Rn  may  be  con- 

Fig.  41  veniently  written  in  the  form 


Rn(x,  a)  =  ^ — pi-  <f> (x,  a). 


EXPANSION   OF   FUNCTIONS  151 

The  problem  is  now  to  determine  <£(#,  a)  so  that  the 
relation 

f(x)=f(a)+f(a)(x-a)+^(x-  af+  .'.. 

+  7 \f,  (x  -  a)H  J  +  yv  '    '  (z  -  ci)n  (1) 

(n  —  1)1  n\ 

may  be  an  algebraic  identity,  in  which  the  right-hand  mem- 
ber contains  only  the  first  n  terms  of  the  series,  with  the 
remainder  after  n  terms.     Thus,  by  transposing, 

f(x)  _/(«)_/(«)(»  -  a)  -fJ£p(x-ay  -  - 

_/->)(a!  _  Bw_»fe«)(a  _  o).=  o.  (2) 

(h-1)!V         ;  n!     ^         ;  w 

Let  a  new  function,  F(z),  be  defined  as  follows: 

F(z)  =f(x)  -/(*)  -f\z){x  -  z)  -£&  (x  -  zf  -  -. 

-O^**-'^1-*^*-*-      (3) 

This  function  F(z)  vanishes  when  z—x,  as  is  seen  by 
inspection,  and  it  also  vanishes  when  z  =  a,  since  it  then 
becomes  identical  with  the  left-hand  member  of  (2) ;  hence, 
by  Rolle's  theorem,  its  derivative  F'(z)  vanishes  for  some 
value  of  z  between  x  and  a,  say  zv     But 

F\z)  =  -/(«)  -|-/(2)  -/»(»)(*  -  »)+/"(*)(»  -«)-... 
(n-l)!V  '      ^(n-l)!1-  ' 


152  DIFFERENTIAL   CALCULUS 

These  terms  cancel  each  other  in  pairs  except  the  last  two ; 
hence  _x 


Since  F'(z)  vanishes  when  z  =  z1}  it  follows  that 

4>(x,  a)  =/<»>&).  (4) 

In  this  expression  zx  lies  between  x  and  a,  and  may  thus  be 

represented  by 

J  Zl  =  a  +  0(x-  a), 

where  6  is  a  positive  proper  fraction.     Hence  from  (4) 
4>(x,  a)=f»[a  +  d(x-a)], 

and  Rn{x,  a)  =/wl>  +  fl(*-«)1  {x _  o)- .• 

The  complete  form  of  the  expansion  of  f(x)  is  then 
/(a>)  =/(«)  +  /»(«) fas-  a)  +^f^  (oc-a)2  +  - 

+f"^  (*-a)»-*+/(WWe(g--«))  («-«)".  (5) 
(n  —  1);  w! 

in  which  n  is  any  positive  integer.  The  series  may  be  carried 
to  any  desired  number  of  terms  by  increasing  n,  and  the  last 
term  in  (5)  gives  the  remainder  (or  error)  after  the  first  n 
terms  of  the  series.  The  symbol  /(n)(«  4-  0  (x  —  a))  indicates 
that  f(x)  is  to  be  differentiated  n  times  with  regard  to  x,  and 
that  x  is  then  to  be  replaced  by  a  +  0(x  —  a). 

*  This  form  of  the  remainder  was  found  by  Lagrange  (1736-1813),  who  pub- 
lished it  in  the  Memoires  de  l'Academie  des  Sciences  a  Berlin,  1772. 


EXPANSION  OF  FUNCTIONS  153 

81.    Another  expression  for  the  remainder.     Instead  of  putting 

RJx,  a)  in  the  form     (x  _  a\» 

r-^fo  a), 

n\ 

it  is  sometimes  convenient  to  write  it 

Rn(x,  a)  =  (x  —  a)  ^(#,  a). 

Proceeding  as  before,  we  find  the  expression  for  F'(z), 

(w-1)! 
In  order  for  this  to  vanish  when  z  =  zlf  it  is  necessary  that 

(n-l)l 
in  which  zt  =  a  +  0(x  —  a),  x  —  zl  =  (x  —  a)(l  —  0). 

Hence     *fe  a)  =/(">(a +  <?(*- a))  (1  _  e).-1(a.  _  a)»-.; 
(n-1)! 

and  !}„(*,  «)=(!-  g)-'/'"y^-a)>(s  -  a)".* 

An  example  of  the  use  of  this  form  of  remainder  is  fur- 
nished by  the  series  for  log  x  in  powers  of  x—  a,  when  x—a 
is  negative,  and  also  in  the  expansion  of  (a  -+-  #)m- 

Ex.  1.  Find  the  interval  of  equivalence  for  the  development  of 
log  x  in  powers  of  x  —  a,  when  a  is  a  positive  number. 

Here,  from  Art.  77,  Ex.  1, 

Xn 

hence  /<»>(a  f  6(x  -a))  =  (-l)n-i       (" >-  1)  » 

(a  +  0(x  —  a))'* 

and,  by  Art.  80, 

^ fl)  =  ^--'c*-*)-  =  (-D-1  r   *-«   r. 

V        '       n(a+0{x-a))n  n        la  +  6(x-a)\ 

*  This  form  of  the  remainder  was  found  by  Cauchy  (1789-1857),  and  first 
published  in  his  "Le<,:ons  sur  le  calcul  infinitesimal,"  1826. 


154  DIFFERENTIAL   CALCULUS 

First  let  x  —  a  be  positive.  Then  when  it  lies  between  0  and  a,  it 
is  numerically  less  than  a  +  0(x  -  a),  since  0  is  a  positive  proper 
fraction  ;  hence  when  n  =  oo 

x~a T  =  0,  and  RJx,  a)  =  0. 

Again,  when  x  —  a  is  negative  and  numerically  less  than  a,  the 
second  form  of  the  remainder  must  be  employed.     As  before 

hence  Rn(x,  a)  =  (1  -  0)«-i .  f"1^"^ 

=  (l_0)»-i.         -(«-')' 


[a-(9(a-.r)]» 

t(a  —  x)  —  0{a  —  x)l  "-1  a  —  x 

a  -0(a  -  x)      J        '  a  -  ${a  -  x)' 

The  factor  within  the  brackets  is  numerically  less  than  1,  hence 
the  (n  —  l)th  power  can  be  made  less  than  any  given  number,  by  tak- 
ing n  large  enough.     This  is  true  for  all  values  of  x  between  0  and  a. 

Therefore,  log  x  and  its  development  in  powers  of  x  —  a  are  equiva- 
lent within  the  interval  of  convergence  of  the  series,  that  is,  for  all 
values  of  x  between  0  and  2  a. 

_i 
Ex.  2.     Show  that  the  development  of  x  ?  in  positive  powers  of 

x  —  a  holds  for  all  values  of  x  that  make  the  series  convergent ;  that 

is,  when,  a;  lies  between  0  and  2  a. 

If  the  function  is  expanded  in  powers  of  x,  the  complete 
form  will  be 

ffljj.        ,  /'-"(0)^-i 


/(*)=/(0)+/'(0)*+-^^+  ...  +^~y;*" 

+  /^M*.  (1) 


EXPANSION  OF  FUNCTIONS  155 

for  the  first  form  of  remainder,  and 

/(*)=/(<>) +/'(0)*  +  /^-'+  ...  +M^-i 

£W(1  _  gy-l  .  x,  (2) 

for  the  second  form  of  remainder. 

Similarly,  the  complete  form  of   Taylor's   series    (Art.  78) 
becomes 


0| 


(n-l)l 


/»>(q  +  flaQ, 


for  the  first  form  of  remainder,  and 


/(a  +  x)  =f(a)  +f(a)x  +  -^z2  +  •  ••  + 


/•"«„2  ,  ,  /'-"(a)  „„-, 


<>   | 


(»-l) 


(4) 


for  the  second  form  of  remainder. 

These  forms  are  of  no  value  for  numerical  computation 
unless  f(n)(x)  can  be  determined,  but  may  sometimes  be  used 
to  advantage  to  obtain  a  maximum  error,  corresponding  to 
small  values  of  n.  It  should  be  observed  that  when  n  =  1, 
the  theorem  of  mean  value  results.     (Art.  39.) 

Ex.  3.  Obtain  the  limit  of  error  in  retaining  but  two  non-vanish- 
ing terms  in  the  expansion  of  log  (x  +  Vl  +  x'1)  when  x  =  \. 


log  (x + viTi"2) = x-  -*. +P-?/-n-?/;r 

wherein  y  =  6x.  LV        J  J  ■ 


24 


156  DIFFERENTIAL   CALCULUS 

The  next  step  is  to  obtain  the  largest  and  the  smallest  value  which 
the  expression  in  brackets  assumes  for  values  of  y  within  the  interval 
0  to  \.     For  this  purpose,  consider  the  function 

„-3.v(3-2.y*) 

(1  + y*)!  ^ 

We  find  that  —  is  positive  for  all  values  of  y  between  y  =  0  and  y  =  \; 
dy 

hence  u  has  its  largest  value  when  y  =  J,  and  the  corresponding  value 
of  the  last  term  is  .000284. 

Ex.  4.  How  many  terms  should  be  used  in  the  expansion  of  ex  in 
powers  of  x  to  insure  a  result  correct  to  four  places  of  decimals  when 
x  =  |? 

Ex.  5.  In  the  expansion  of  logio  (1  +  x)  in  powers  of  x  how  many 
terms  should  be  used  in  order  to  obtain  the  value  of  logio  (1  •  8)  cor- 
rect to  5  decimals  ? 


CHAPTER   XI 

INDETERMINATE    FORMS 

82.  Hitherto  the  values  of  a  given  function  f(x),  corresponding 
to  assigned  values  of  the  variable  x,  have  been  obtained  by  direct 
substitution.  The  function  may,  however,  involve  the  variable 
in  such  a  way  that  for  certain  values  of  the  latter  the  correspond- 
ing values  of  the  function  cannot  be  found  by  mere  substitution. 

For  example,  the  function 


sin  x 


for  the  value  x  =  0,  assumes  the  form  -,  and  the  correspond- 
ing value  of  the  function  is  thus  not  directly  determined.  In 
such  a  case  the  expression  for  the  function  is  said  to  assume 
an  indeterminate  form  for  the  assigned  value  of  the  variable. 

The  example  just  given  illustrates  the  indeterminateness  of 
most  frequent  occurrence;  namely,  that  in  which  the  given 
function  is  the  quotient  of  two  other  functions  that  vanish  for 
the  same  value  of  the  variable. 

Thus  if  f(x)  =  *M, 

and  if,  when  x  takes  the  special  value  a,  the  functions  <f>(x) 
and  \p(x)  both  vanish,  then 

J  K   '      *(a)      0 
is  indeterminate  in  form,  and  cannot  be  rendered  determinate 
without  further  transformation. 

157 


158 


DIFFERENTIAL   CALCULUS 


83.    Indeterminate   forms    may   have   determinate   values.     A 

case  has  already  been  noticed  (Art.  2)  in  which  an  expression 
that  assumes  the  form  -  for  a  certain  value  of  its  variable  takes 

a  definite  value,  dependent  upon  the  law  of  variation  of  the 
function  in  the  vicinity  of  the  assigned  value  of  the  variable. 
As  another  example,  consider  the  function 

x2  —  a2 


y  = 


x—  a 


If  this  relation  between  x  and  y  is  written  in  the  forms 

y(x  —  a)=x2  —  a2,     (x  —  a)(y  —  x  —  a)  =  0, 

it  will  be  seen  that  it  can  be  represented  graphically,  as  in  Fig. 
42,  by  the  pair  of  lines 

x  —  a  =  0, 
y  —  x  —  a  =  0. 

Hence  when  x  has  the  value  of  a,  there 
is  an  indefinite  number  of  corresponding 
points  on  the  locus,  all  situated  on  the 
line  x  =  a;  and  accordingly  for  this 
value  of  x  the  function  y  may  have  any 
value  whatever,  and  is  therefore  indeterminate. 

When  x  has  any  value  different  from  a,  the  corresponding 
value  of  y  is  determined  from  the  equation  y  =  x-\-a.  Now, 
of  the  infinite  number  of  different  values  of  y  corresponding 
to  x  =  a,  there  is  one  particular  value  AP  which  is  continuous 
with  the  series  of  values  taken  by  y  when  x  takes  successive 
values  in  the  vicinity  of  x  =  a.  This  may  be  called  the  de- 
terminate value  of  y  when  x  —  a.  It  is  obtained  by  putting 
x  =  a  in  the  equation  y  =  x  +  a,  and  is  therefore  y  =  2  a. 


Fig.  42 


INDETERMINATE   FORMS  159 

This  result  may  be  stated  without  reference  to  a  locus  as 
follows.    When  x  =  a,  the  function 


x—  a 

is  indeterminate,  and  has  an  infinite  number  of  different 
values ;  but  among  these  values  there  is  one  determinate  value 
which  is  continuous  with  the  series  of  values  taken  by  the 
function  as  x  increases  through  the  value  a ;  this  determinate 
or  singular  value  may  then  be  defined  by 

liui  xr—  a2^ 
x=a  x  —  a 

In  evaluating  this  limit  the  common  factor  x—  a  may  be  re- 
moved from  numerator  and  denominator,  since  this  factor  is 
not  zero  while  x  is  different  from  a;  hence  the  determinate 
value  of  the  function  is 

lim  a?  +  «_o„ 

Ex.  1.     Find  the  determinate  value,  when  x  =  1,  of  the  function 

x3  +  2  x2  -3x 
3  xb  _  3  xa  _  x  +  i ' 

which,  at  the  limit,  takes  the  form  — 

0 

This  expression  may  be  written  in  the  form 

(x*  +  Zx)(z-  1) 
(3s3  -  1)0-  1)' 

which  reduces  to  x  +  '  x •      When  x  =  1,  this  becomes  -  =  2. 
3  x2  -  1  2 

Ex.  2.     Evaluate  the  expression 

x8  +  ax2  +  a2x  4-  a3 


,  x8  4-  b2x  +  az2  +  a&2 

when  x  =  —  a. 


160  DIFFERENTIAL   CALCULUS 

Ex.  3.   Determine  the  value  of 

x8  -  7  x2  +  3  x  +  14 

x3  +  3  x2  -  17  x  +  14 
when  x  =  2. 


Ex.  4.   Evaluate     - — — —  when  x  =  0. 


(Multiply  both  numerator  and  denominator  by  a  -f  Va2  —  x2.) 

84.    Evaluation  by  development.     In  some  cases  the  common 
vanishing  factor  can  be  best  removed  after  expansion  in  series. 

Ex.  1.   Consider  the  function  mentioned  in  Art.  82, 

e*  -  e~* 
sin  x 

When  numerator  and  denominator  are  developed  in  powers  of  x, 
the  expression  becomes 

21      31^  V  21      31  / 

*-£+... 
3! 

2,  +  ^+...      2+f+... 


x3    ,  -        X2    . 

—  ST+-      1-«  +  - 

which  has  the  determinate  value  2,  when  x  takes  the  value  zero. 

Ex.  2.     As  another  example,  evaluate,  when  x=  0,  the  function 

x  —  sin-1  a: 
sin8x 

By  development  it  becomes 


«+ 


XJ  + 


Removing  the  common  factor,  and  then  putting  x  =  0,  we  obtain 


INDETERMINATE    FORMS  161 

In  these  two  examples  the  assigned  value  of  x,  for  which  the 
indeterininateness  occurs,  is  zero,  and  the  developments  are  made 
in  powers  of  x.  If  the  assigned  value  of  x  is  some  other  number, 
as  a,  then  the  development  should  be  made  in  powers  of  x  —  a. 

Ex.  3.   Evaluate,  when  x  =  -,  the  function 
2 

COS  X 


1  —  sin  x 

IT 


By  putting  x  —  -  =  h,  x  =  -  +  h,  the  expression  becomes 

eo.(=+»)  .  -»+£-...      _l+£ 

\  2        /  —  sm  7i  6  6 


.        .    hr  ,    ,\       1  -cos A       A2      A4  ,  /i      A8   , 

1  —  sin    — \-  h)  — h  •••  h  ••• 

\2         /  2      24  2      24^ 

which  becomes  infinite  when  7*  =  0 ;  that  is,  when  x  =  -• 

2 

TJ  lim      cos  x 

Hence  .  m  - =  ±  oo, 

x=fl-siuo:  ' 

according  as  h  approaches  zero  from  the  negative  or  positive  side. 

85.    Evaluation   by  Differentiation.     Let   the  given   function 
be  of  the  form  ^~,  and  suppose  that  /(a)  =  0,  cf>(a)  =  0.     It 

is  required  to  find     ^  ^-L-i  . 
4  x=a<f>(x) 

We  assume  that  f(x),  <j>(x)  can  be  developed  in  the  vicinity 

of  x  =  a,  by  expanding  them  in  powers  of  x  —  a.     Then 

m  /(«)+/'(«)(*  -  «)+-^(f  (»-  «)2  +  - 

*(*)      <l>(a)  +  4>'(a)(x  -  a)  +  *!M (x  -  uy+  - 
/•(«)(a(_a)+^2l((,_«)i+... 


^«)(*-a)  +  ^(z-a)2+... 

EL.    CALC.  11 


162  DIFFERENTIAL   CALCULUS 

By  dividing  by  x  —  a  and  then  letting  x  =  a,  we  obtain 

lim  f(x)_f(a) 
«-«  +  («)      +'(o) 

By  hypothesis  the  functions  f'(a),  <£'(a)  will  both  be  finite. 

If /'(a)  =  0,  <£'(a)  =£  0,  then  £i&  =  0. 

<j>(a) 

lf/'(a)*0,  *'(a)  =0,  then  ^  =  oo. 

<£(a) 

If  /'(a)  and  <£'(a)  are  both  zero,  the  limiting  value  of  J-^-l 

is  to  be  obtained  by  carrying  Taylor's  development  one  term 

farther,  removing  the  common  factor  (x  —  a)2,  and  then  letting 

f"(a) 
x  approach  a.     The  result  is  J—~  • 

Similarly,  if /(a), /'(a),  /"(a);  </>(a),  <^)'(a),  <^>"  (a)  all  vanish, 
it  is  proved  in  the  same  manner  that 

lim   f(x)  =  f'"(a) 
x  =  a'<j>(x)      <£'"(«)' 
and  so  on,  until  a  result  is  obtained  that  is  not  indeterminate 
in  form. 

Hence  the  rule : 

To  evaluate  an  expression  of  the  form  -,  differentiate  numer- 
ator and  denominator  separately ;  substitute  the  critical  value  of 
x  in  their  derivatives,  and  equate  the  quotient  of  the  derivatives 
to  the  indeterminate  form. 

Ex.  1.     Evaluate  *  ~  T  ^  when  0  =  0. 

Put  f(0)  =  1  -  cos  6,       <f>(0)  =  e*. 

Then  f(0)  =  sin  0,  <f>'(0)=2$t 

and  /'(0)  -  0,  <£'(0)  =  0. 


INDETERMINATE   FORMS  163 

Thus,  the  function  \  *  J    is  also  indeterminate  at  0  =  0.     It  is  there- 
9  (0) 

f"({y\ 
fore  necessary  to  obtain  </,,„  '  • 

Accordingly,     f"{6)  =  cos  0,  p"(0)  =  2, 

/'(0)=1,  9"(0)=2, 

hence  lim  1  —  cos  0  _  1  J 

0  =  0       02       -2* 

Ex.2.     Findx=,0 

lim    i*  +  e~*  +  2  cos  a:  —  4        lim    ex  —  e~x  —  2  sin  a: 


x  =  0 

4  a:3 

lim  e*  +  e 

*  —  2 

COS  X 

x=0 

12  a:2 

lim   ex  —  e 

-*  +  2 

sin  x 

x  =  0 

24  x 

lim  ez  +  e 

-*  +  2 

cosx 

x  =  0  24 

1 
6' 


Ex.  3.   Find     lim  a;-siD3rcosa:. 
x  =  0  x8 


Ex.4.   Find    M"  «,-2»'i**+»«-*. 

*  =  1        a:4  -  2  a:3  +  2  a:  -  1 

In  this  example,  show  that  z  —  1  is  a  factor  of  both  numerator  and 

denominator. 

^      _     „.    ,     lim   3  tan  x  -  3  a:  -  a:8 
Ex.  5.    Find  „  .  n 

In  applying  this  process  to  particular  problems,  the  work 
can  often  be  shortened  by  evaluating  a  non-vanishing  factor 
in  either  numerator  or  denominator  before  performing  the 
differentiation. 


164  DIFFERENTIAL   CALCULUS 

Ex.  6.   Find    ";■>   (»-*)' tan*. 

X  ==  "  X 

The  given  expression  may  be  written 

lim  .     tanx_    lim  lim    tans 

=  16-1  =  16. 

In  general,  if  f(x)  =  *P(%)x(x)>   an(^  ^    «M°0  =  0>  x(a)^0, 
(h (a)  =  0,  then  ,.        .,  N  ,,,  * 

llm  /M  =  vr^^M. 

*  =  «*(*)      XU^>'(a) 
For 

lim    iKx)x(x)=   lim      ,  v  .     lim    </<»)  a)  .  ^!M 

x  =  a      $(x)  x  =  a^K  >     x  =  a^{x)      *     ;     <f>'(a) 

lim    sin  x  cos2  x 


Ex.  7.     Find 


x  =  $  (2  x  -  tt)2 


Ex.8.    Find    lim    (.-3)2log(2      x}> 
x  =  1  sin(x  -  1) 


EXERCISES 

Evaluate  the  following  expressions  : 

1.    1-CO8arwhenx  =  0.  7.    e*  +  e~'  ~  2  when  x  =  0. 


2.    e*  .    e       when  a;  =  0.  8     tan  a;- sin  x  cos  a:  when  ^_a 


3. 

*3  -  1        U                  1 

when  x  =  1. 

x-  1 

4. 

when  x  =  0. 

&*  -  1 

5. 

sin  ax      ,               „ 

when  x  =  0. 

sin  frx 

l 

9.    when  x  =  0. 

tan-1  a; 


10.    g*B"i*-*-*    when  a- =  0. 
x2  +  a;log  (1-a:) 

>.     (1  4- arV»  —  1       ,  n        -.-.      tan  a;  —  sin  a;      ,  n 

6.    Vx  ^  x' 1  when  x  =  0.       11.    -Q when  x  =  0. 

x  Xs 

There  are   other   indeterminate   forms  than  -•     They   are 
£,00-00,0°,  r,  00°. 


INDETERMINATE   FORMS 


165 


86.    Evaluation  of  the  indeterminate  form  ® . 

GO 

fix) 
Let  the    function  J  \  '    become   5°-   when  x  =  a.     It   is  re- 
<j>(x)  <*> 

quired  to  find    lim  &&• 

This  function  can  be  written 


<K*)      _1_' 

which   takes   the   form  -   when  x  =  a,  and  can  therefore  be 

evaluated  by  the  preceding  rule. 
When  x  =  a, 


—   vv, 

lim    /(»)  _   lim 
x  =  a  <£(#)      £  =  a 

1                   4,\x) 

*(*)_    lim         [*(*)]* 

1         ic  =  a         f'(x) 

lim 

~" "  x  —  a 

-/W  T+'(*). 

(1) 


If  both  members  are  divided  by      im  -£-^J   when  this  limit 

is  not  0  nor  co,  the  equation  becomes 

1=    lim   f(x)  +\x) 
*±<*<f>(x)f'(x)' 


therefore 


lim 

x  =  a 


d,{x) 


=/'(<0 
+'(«) 


(2) 


This  is  exactly  the  same  result  as  was  obtained  for. the  form 
;  hence  the  procedure  for  evaluating  the  indeterminate  forms 


-,  g,  is  the  same  in  both  cases. 
0    ^ 


166  DIFFERENTIAL   CALCULUS 

When  the  true  value  of  ■->--'  is  0  or  go  ,  equation  (1)  is  satis- 
fied,  independent  of  the  value  of      ,~ ;  but  (2)  still  gives  the 

correct    value.     For,    suppose       ™  ^  w  =  0.     Consider    the 
function 

which  has  the  form  °°-  when  a;  =  a,  and  has  the  determinate 

GO  7 

value  c,  which  is  not  zero.     Hence  by  (2) 

lim  fix)  +  c<f>  (x)  =  /'  (a)  +  c<f>'  (a)  =  /'  (a)       c 
*  =  a         <f>(x)  4>'(a)  <j>'  (a) 

Therefore,  by  subtracting  c, 

lim    /(a?)  =/(a)  , 
•   £  =  «<£(&)       </>'(a) 

If    "F-fiS.  =00,  then    "™  ^  =  0,  which  can  be  treated 
x~a<f>(x)  x-af(x) 

as  the  previous  case. 

The  forms  0  •  go  and  go-go  can  usually  be  evaluated  by 
putting  them  in  one  or  the  other  of  the  forms  already  dis- 
cussed. In  the  case  of  the  others,  in  which  the  indetermi- 
nateness  appears  in  the  exponent,  the  logarithm  of  the 
function  can  be  reduced  to  one  of  the  preceding  forms. 

EXERCISES 

Evaluate  the  following  expressions: 
1.   logsi»2*  when  x  =  0. 


log  sin  x 
2.   !2££  ^ 


4. 

tan  x 
tan  5  x 

when 

x 

_  IT 

5. 

sec  3  a: 

when 

X 

_  IT 

cot  x  sec  5  x  2 

6.  xs[nx  when  x  =  0. 

7.  (cos  ax)csc2cx  when  x  =  0. 


^n  6.    xB[QX  when  x  =  0. 

3.    —  when  x  =  go. 


IT) 


CHAPTER   XII 

CONTACT  AND   CURVATURE 
87.    Order  of  contact.     The  points  of  intersection  of  the  two 

are  found  by  making  the  two  equations  simultaneous ;  that  is, 
by  finding  those  values  of  x  for  which 

Suppose  x  =  a  is  one  v?lue  that  satisfies  this  equation. 
Then  the  point  x  =  a,  y  =  A(i)  =  if; (a)  is  common  to  the  curves. 

If,  moreover,  the  two  curves  have  the  same  tangent  at  this 
point,  they  are  said  to  touch  each  other,  or  to   have  contact 

with   each   other.     The  values  of   y  and  of  —   are  thus  the 

J  dx 

same  for  both  curves  at  the  point  in  question,  which  requires 

that  *(«)=</<(«), 

<j,\a)=4,\a). 

d2v 
If,  in  addition,  the  value  of  — "  is  the  same  for  each  curve 

dx~ 
at  the  point,  then 

and  the  curves  are  said  to  have  a  contact  of  the  second  order 
with  each  other,  provided  cf>'"(a)  =£  i//"(a). 

If  <f>(a)  =\f/(a),  and  all  the  derivatives  up  to  the  nth  order 
inclusive  are  equal  to  each  other,  that  is,  <f>'(a)  =  if/' (a), 
<}>"(a)  =  i(,"(a),  .-.,  cf>(n\a)  =<A(n)(a),  but  <f>(n+1)(a)  =£  fn+1)(a), 
the  curves  are  said  to  have  contact  of  the  ?ith  order. 

167 


168  DIFFERENTIAL   CALCULUS 

88.  Number  of  conditions  implied  by  contact.  or  j  equation 
of  the  curve  y  =  <f>(x)  is  given,  and  it  is  requir  ,.  to  determine 
the  equation  of  a  second  curve  y  =  \\j  (x)  that  shall  have  contact 
of  any  given  order  with  y  =  <f>(x)  at  a  specified  point,  then, 
from  the  definition  given  in  the  preceding  article  for  contact 
of  the  nth  order,  n  -f  1  conditions  must  be  imposed  upon  the 
coefficients  in  \p(x).  The  required  curve  must  therefore  con- 
tain at  least  n  -f- 1  arbitrary  constants  in  >rder  to  fulfill  the 
required  conditions. 

A  straight  line  has  two  arbitrary  constants,  which  can  be 
determined  by  two  conditions ;  accordingly  a  straight  line  can 
be  drawn  which  touches  a  given  curve  at  any  specified  point. 
For  if  the  equation  of  a  line  is  written  y  =  mx  -f  b,  then 

dy  dr.;     p. 

dx  dx 

hence,  through  any  arbitrary  point  x  =  a  on  a  given  curve 
y=cf>(x),  a  line  can  be  drawn  which  has  contact  of  the  first 
order  with  the  curve,  but  which  has  not  in  general  contact  of 
the  second  order;  for  the  two  conditions  for  first-order  con- 
tact are  ma  +  6  =  <£(«), 

m  =  <£'(«), 
which  are  just  sufficient  to  determine  m  and  b. 

In  general  no  line  can  be  drawn  having  contact  of  an  order 
higher  than  the  first  with  a  given  curve  at  a  given  point ;  but 
there  are  certain  special  points  at  which  this  can  be  done. 
For  example,  the  additional  condition  for  second-order  contact 
is  0  =  <f>"(a),  which  is  satisfied  when  the  point  x  =  a  is  a 
point  of  inflexion  on  the  given  curve  y=<j>(x).  (Art.  49.) 
Thus  the  tangent  at  a  point  of  inflexion  on  a  curve  has  contact 
of  the  second  order  with  the  curve. 


CONTACT   AND   CURVATURE  169 

The  equation  of  a  circle  has  three  independent  constants. 
It  is  therefore  possible  to  determine  a  circle  having  contact  of 
the  second  order  with  a  given  curve  at  any  assigned  point. 

The  equation  of  a  parabola  has  four  constants,  hence  a 
parabola  can  be  found  which  has  contact  of  the  third  order 
with  the  given  curve  at  a  given  point. 

The  general  equation  of  a  central  conic  has  five  independent 
constants,  hence  a  conic  can  be  found  which  has  contact  of  the 
fourth  order  with  a  given  curve  at  any  specified  point. 

As  in  the  case  of  the  tangent  line,  special  points  may  be 
found  for  which  these  curves  have  contact  of  higher  order. 

89.    Contact  of  odd  and  of  even  order. 

Theorem.  At  a  point  where  two  curves  have  contact  of  an 
odd  order  they  do  not  cross  each  other;  but  they  do  cross 
where  they  have  contact  of  an  even  order. 

For,  let  the  curves  y  =  <f>(x),  y=\f/(x)  have  contact  of  the  ?ith 
order  at  the  point  whose  abscissa  is  a ;  and  let  yx,  y2  be  the 
ordinates  of  these  curves  at  the  point  whose  abscissa  is  a  +  h. 

Then  y1  =  cf>(a  +  h),   y2=z  ij,(a-\-h), 

and  by  Taylor's  theorem 

Vl  =  *(a)  +  *'(a)  •  h  +  *^)  •  ft2  +  •  •  • 

+  *^.».+     *^«(a)+.... 

n !  (?i  -f  1) ! 


fc.^(o)+^(a).*+*^i.V 


+  • 


n !  (n  +  1)  ! 


170  DIFFERENTIAL   CALCULUS 

Since  by  hypothesis  the  two  curves  have  contact  of  the  nth 
order  at  the  point  whose  abscissa  is  a,  hence 

4(a)  =$(a),  <j>'(a)=ip'  a),  ..'.,  <y(a)  =  ^(a), 
and         yi-y^-J^ir+1(a)+  ...  _r+i(a)_  ...]; 

but  this  expression,  when  h  is  sufficiently  diminished,  has  the 
same  sign  as  .,,,r  „,.>  , , ,  v _ 

Hence,  if  n  is  odd,  yl  —  y2  does  not  change  sign  when  h  is 
changed  into  —  h,  and  thus  the  two  curves  do  not  cross  each 
other  at  the  common  point.  On  the  other  hand,  if  n  is  even, 
ll\  —  V*  changes  sign  with  7i ;  and  therefore  when  the  contact 
is  of  even  order  the  curves  cross  each  other  at  their  common 
point. 

Geometrically,  we  may  say  that  two  curves  having  contact 
of  the  nth  order  pass  through  n  -\-  1  common  points  which 
approach  coincidence  at  the  point  of  contact.  For  let  y  =  <f>(x), 
y  =  xf/(x)  touch  each  other  at  x  =  a.  This  means  that  they  have 
two  coincident  points  in  common  at  (a,  <f>(a)),  and  the  condi- 
tions to  be  satisfied  are 

<f>(a)=^(a),    *'(a)  =  *'(a). 

If  the  curves  also  have  a  point  in  common  for  x  =  a  -f-  h,  then 

<f>(a  +  h)=ip(a  +  h). 

Expanding  by  Taylor's  series  and  making  use  of  the  preced- 
ing conditions,  we  may  cancel  the  common  factor  Jr.  If  now 
this  condition  is  still  satisfied  when  h  approaches  zero,  so  that 
the  third  point  of  intersection  approaches  the  position  of  the 
two  coincident  ones,  then  we  must  have  the  further  condition 


CONTACT   AND   CURVATURE  171 

<£"(a)=  \J/"(a).  Thus,  three  coincident  points  of-  intersection 
imply  contact  of  the  second  order.  By  repeating  this  argu- 
ment the  above  theorem  results. 

For  example,  the  tangent  line  usually  lies  entirely  on  one 
side  of  the  curve,  but  at  a  point  of  inflexion  the  tangent  crosses 
the  curve. 

Again,  the  circle  of  second-order  contact  crosses  the  curve 
except  at  the  special  points  noted  later,  in  which  the  circle 
has  contact  of  the  third  order. 

EXERCISES 

1.  Find  the  order  of  contact  of  the  curves 

4  y  =  x2  and  y  =  x  —  1. 

2.  Find  the  order  of  contact  of  the  curves 

x  =  ys  and  x  —  2  y  +  1  =  0. 

3.  Find  the  order  of  contact  of  the  curves 

4  y  =  x2  —  4  and  x2  —  2  y  —  3  —  y2. 

4.  Determine  the  parabola  having  its  axis  parallel  to  the  #-axis, 
which  has  the  closest  possible  contact  with  the  curve  a2y  —  x3  at  the 
point  («,  a).  (The  equation  of  a  parabola  having  its  axis  parallel 
to  the  y-axis  is  of  the  form 

y  =  Ax2  +  Bx  +  C.) 

5.  Determine  a  straight  line  which  has  contact  of  the  second  order 
with  the  curve  y  =  x8  -  3x2  -  9  x  +  9. 

6.  Find  the  order  of  contact  of 

y  =  log(x  —  1)  and  x2  —  6x  +  2y  +  8  =  0 
at  the  point  (2,  0). 

7.  What  must  be  the  value  of  a  in  order  that  the  curves 

y=x+l-\-a(x—  l)2  and  xy  =  3  x  —  1 
may  have  contact  of  the  second  order? 


172  DIFFERENTIAL   CALCULUS 

8.  Determine  the  parabola  which  has  its  axis  parallel  to  the  y-axis 
and  has  contact  of  the  second  order  with  the  hyperbola  xy  =  1  at  the 
point  (1,  1). 

9.  Determine  the  point  and  order  of  contact  of  the  curves 

(a)  y=x*,  y  =  6  x2  -  9  x  +  4 ; 
(6)  y  =  xs,  y  =  -Qx2-12x-8. 

10.  Determine  the  parabola  which  has  its  axis  parallel  to  the  y-axis, 
passes  through  the  point  (0,  3),  and  has  contact  of  the  first  order 
with  the  curve  y  =  2  x2  at  the  point  (1,  2).  Similarly  for  a  parabola 
having  its  axis  parallel  to  the  x-axis. 

11.  Show  that  the  curve  y  =  sin  x  has  contact  of  the  sixth  order 

with  the  curve  xz  ,    x5 

v  =  x 

*  6       120 

at  the  origin.     Show  that  y  =  sin  x,  y  =  sinh  x,  have  contact  of  the 

second  order  at  the  origin.     Draw  these  curves. 

12.  Find  the  order  of  contact  of  the  curves  y  =  cos  x,  y  =  cosh  x 
at  the  point  (0,  1).     Sketch  the  curves. 

13.  Find  the  order  of  contact  at  the  origin  of  the  curves 

,  ■  ,      ,      _  sinh  x 

y  =  tan  x,  y  =  tanh  x  = . 

cosh  x 

90.  Circle  of  curvature.  The  circle  that  has  contact  of  the 
closest  order  with  a  given  curve  at  a  specified  point  is  called 
the  osculating  circle  or  circle  of  curvature  of  the  curve  at  the 
given  point.  The  radius  of  this  circle  is  called  the  radius  of 
curvature,  and  its  center  is  called  the  center  of  curvature  at  the 
assigned  point. 

91.  Length  of  radius  of  curvature;  coordinates  of  center  of 
curvature.     Let  the  equation  of  a  circle  be 

(X-ay  +  (Y-pf  =  R\  (1) 

in  which  R  is  the  radius,  and  a,  ft  are  the  coordinates  of  the 
center,  the  current  coordinates  being  denoted  by  X,  Yto  dis- 


CONTACT   AND   CURVATURE 


173 


tinguish  them  from  the  coordinates  of  a  point  on  the  given 
curve. 

It  is  required  to  determine  R,  a,  /?,  so  that  this  circle  may 
have  contact  of  the  second  order  with  the  given  curve  at  the 
point  (x,  y). 

From  (1),  by  successive  differentiation,  we  deduce 


(X-o)  +  (F-/8)g=0, 

\dX        K        H'dX2 


(2) 


If  the  circle  (1)  has  contact  of  the  second  order  at  the  point 
(x,  y)  with  the  given  curve,  then  when  X  =  x  it  is  necessary 
that 


dY=c]y    d2Y=d2y 
dX     dx    dX2     dx2' 


(3) 


Substituting  these  expressions  in  (2),  we  obtain 


dy 


(x-  a)  +(3,-^)^  =  0, 


whence 


y-P- 


H% 


dx2 


dy 

dx 
x~  a=  — 


L+(D" 

<Py 

dx* 


(4) 


(5) 


and  finally,  by  substitution  in  (1), 


R 


.MIT) 


dx2 


(6) 


174  DIFFERENTIAL   CALCULUS 


Ex.  1.     For  the  curve  y  =  sin  x,  show  that  a  =  x-\-  cot  x(l  +  cos2  x), 

s 
j3  =  —  2  cos  a:  esc  x,    it  =  —  (1  +  cos'2  x)2  esc  x.     Find   the   numerical 

values  of  a  and  B  when  a:  =  0,  — ,  — ,  — ,  and  locate  the  corresponding 
r  6     3     2 

points  («,  /3)  on  a  drawing.     Sketch  roughly  the  path  of  this  point 
as  x  varies.     Write  the  equation  of  the  osculating  circle  for  the  point 

x  =  -  ,  and  for  x  =  -.     Draw  these  circles. 
3  2 

Ex.  2.  For  the  curve  y  =  xs,  find  a,  /?,  R  in  terms  of  x.  Compute 
their  numerical  values  at  x  —  1,  .7,  .5,  .3,  0.  Show  that  i?  is  a  mini- 
mum when  x  —  — —  =  .39  •••,  and  that  the  value  of  R  is  .57  •••• 

92.  Limiting  intersection  of  normals.  Let  P=  (#1?  y-^)  and 
P'  =  (a?2,  y2)  be  Wo  points  on  a  given  curve  f(x,  y)  =  0.  The 
equations  of  the  normals  at  these  points  are 

(a._a.2)  +  (2/_2/2)^?  =  0< 

d#2 

If  («',  /?')  is  the  point  of  intersection  of  these  two  lines,  then 

dxx 
(^_iK2)  +  (/3'_2/2)^==0. 

Now  consider  the  function  if/(x)  of  a:  denned  by  the  equations 
f(x)  =  (x-  a')  +  (y-  /?')  £,        f(x,  y)  =  0. 

Since   i/a(^1)  =  0   and   ^(#2)  =  0,  hence   by   Rolle's   theorem 

(Art.  79)  it  follows  that 

t'(x)  =  0, 


CONTACT  AND  CURVATURE  175 

in  which  x  is  defined  by  the  inequalities 

U/j  <^,  X  <C  X2> 

Hence  a',  ft'  may  be  determined  by  the  simultaneous 
equations  ^(^)  =  0,    f(5)  =  0. 

When  P'  =  (x2,  y2)  approaches  coincidence  with  the  point 
P=(xx,  yx),  then  if/'  (x)  =  i//  (a;,),  and  therefore  from  (4),  the 
point  («',  /8')  becomes  the  center  of  curvature,  hence : 

The  center  of  curvature  at  a  point  Pon  a  curve  is  the  limiting 
position  of  the  point  of  intersection  of  the  normal  at  P  with  the 
normal  at  the  point  P',  when  P'  approaches  the  position  of  P. 

93.    Direction  of  radius  of  curvature.     Since,  at  any  point  P 

on  the  given  curve,  the  value  of  -*  is  the  same  for  the  curve 

dx 

and  the  osculating  circle  for  that  point,  it  follows  that  they 
have  the  same  tangent  and  normal  at  P,  and  hence  that  the 
radius  of  curvature  coincides  with  the  normal.     Again,  since 

the  value  of  —  is  the  same  for  both  curves  at  P,  it  follows 
dx2 

from  Art.  50,  that  they  have  the  same  direction  of  bending 
at  that  point,  and  hence  that  the  center  of  curvature  lies  on 
the  concave  side  of  the  given  curve  (Fig.  43). 

It  follows  from  this  fact  and  Art.  87  that  the  osculating 
circle  is  the  limiting  position  of  a  circle  passing  through  three 
points  on  the  curve  when  these  points  move  into  coincidence. 

The  radius  of  curvature  is  usually  regarded  as  positive  or 
negative  according  as  the  bending  of  the  curve  is  positive  or 

d2v 
negative  (Art.  49),  that  is,  according  as  the  value  of    — -~  is 

ctx~ 
positive  or  negative ;  hence,  in  the  expression  for  R,  the  radi- 
cal in  the  numerator  is  always  to  be  given  the  positive  sign. 


176 


DIFFERENTIAL   CALCULUS 


The  sign  of  R  changes  as  the  point  P  passes  through  a  point 
of  inflexion  on  the  given  curve  (Fig.  44).  It  is  evident  from 
the  figure  that  in  this  case  R  passes  through  an  infinite  value ; 


Fig.  43  Fig.  44 

for  the  circle  through  the  points  N,  P,  Q  approaches  coinci- 
dence with  the  inflexional  tangent  when  N  and  Q  approach 
coincidence  with  P,  and  the  center  of  this  circle  at  the  same 
time  passes  to  infinity. 

94.  Total  curvature  of  a  given  arc ;  average  curvature.  The 
total  curvature  of  an  arc  PQ  (Fig.  45)  in  which  the  bending 
is  in  one  direction,  is  the  angle  through 
which  the  tangent  swings  as  the  point 
of  contact  moves  from  the  initial  point 
P  to  the  terminal  point  Q ;  or,  in  other 
words,  it  is  the  angle  between  the  tan- 
gents at  P  and  Q,  measured  from  the 
former  to  the  latter.  Thus  the  total 
curvature  of  a  given  arc  is  positive  or  negative  according  as 
the  bending  is  in  the  positive  or  negative  direction  of  rotation. 
The  total  curvature  of  an  arc  divided  by  the  length  of  the 
arc  is  called  the  average  curvature  of  the  arc.     If  the  length  of 


Fig.  45 


CONTACT  AND   CURVATURE  177 

the  arc  PQ  is  As  centimeters,  and  if  its  total  curvature  is  A<£ 

radians,  then  its  average  curvature  is  — —  radians  per  centimeter. 

As 

95.   Measure  of  curvature  at  a  given  point.     The  measure  of 

the  curvature  of  a  given  curve  at  a  given  point  P  is  the  limit 

which  the  average  curvature  of  the  arc  PQ  approaches  when 

the  point  Q  approaches  coincidence  with  P. 

Since   the   average   curvature    of   the   arc   PQ  is  — — ,   the 

As 
measure  of  the  curvature  at  the  point  P  is 

lim    A<£  _  dcf> 


As=0  As       d 


and  may  be  regarded  as  the  rate  of  deflection  of  the  arc  from 
the  tangent  estimated  per  unit  of  length ;  or,  as  the  ratio  of 
the  angular  velocity  of  the  tangent  to  the  linear  velocity  of  the 
point  of  contact. 

To  express  k  in  terms  of  x,  y,  and  the  derivatives  of  y,  we 

observe  that 

whence 

and 


tan  <£  = 

dx ' 

*  = 

dx 

d<f>_ 
ds 

ds  \          dxj 

dx\          dxj   ds 

<l2y 

dx2         1 

\dxj    dx 

cPy 

defy 

dx2 

K~ds  = 

Mtw 

therefore  k  =  ^  = "^7     9       •  [Art.  41. 

EL.   CALC.  — 12 


178  DIFFERENTIAL   CALCULUS, 

96.  Curvature  of  an  arc  of  a  circle.  In  the  case  of  a  circular 
arc  the  normals  are  radii ;  * 

hence  As  =  r.A0,   ^  =  -,  (1) 

As      r 

and  therefore  k  =  -. 

r 

It  follows  that  the  average  curvature  of  all  arcs  of  the  same 

circle  is  constant  and  equal  to  -  radians  per  unit  of  length. 

r 

For  example,  in  a  circle  of  2  feet  radius  the  total  curvature 
of  an  arc  of  3  feet  is  §  =  1.5  radians,  and  the  average  curva- 
ture is  .5  radian  per  foot. 

It  also  follows  from  (1)  that  in  different  circles,  arcs  of  the 
same  length  have  a  total  curvature  inversely  proportional  to 
their  radii. 

Thus  on  a  circumference  of  1  meter  radius,  an  arc  of  5  decimeters 
has  a  total  curvature  of  .5  radian,  and  an  average  curvature  of  .1 
radian  per  decimeter  ;  but  on  a  circumference  of  half  a  meter  radius, 
the  same  length  of  arc  has  a  total  curvature  of  1  radian  and  an 
average  curvature  of  .2  radian  per  decimeter. 

97.  Curvature  of  osculating  circle.  A  curve  and  its  osculat- 
ing circle  at  P  have  the  same  measure  of  curvature  at  that 
point. 

For,  let  k,  k'  be  their  respective  measures  of  curvature  at 
the  point  of  contact  (x,  y).     Then  from  Art.  95, 

&y 

dx2 


\>+m\ 


CONTACT   AND   CURVATURE  179 

But  this  is  the  reciprocal  of  the  expression  for  the  radius  of 
curvature  (Eq.  (6),  p.  173) ;  hence 

1 

K  =  • 

R 

That  is  :  the  measure  of  curvature  k  at  a  point  P  is  the  recipro- 
cal of  the  radius  of  curvature  R  for  that  point.  Since  a  curve 
and  its  osculating  circle  have  the  same  radius  of  curvature 
(Art.  90)  at  their  point  of  contact,  it  follows  from  this  result 
that  the  measure  of  curvature  is  also  the  same  for  both;  *=*'. 

It  is  on  account  of  this  property  that  the  osculating  circle 
is  called  the  circle  of  curvature.  This  is  sometimes  used  as 
the  denning  property  of  the  circle  of  curvature.  The  radius 
of  curvature  at  P  would  then  be  denned  as  the  radius  of  the 
circle  whose  measure  of  curvature  is  the  same  as  that  of  the 
given  curve  at  the  point  P.  Its  value,  as  found  from  Art.  95 
and  Art.  96,  accords  with  that  given  in  Art.  91. 

EXERCISES 

1.  Find  the  radius  of  curvature  of  the  curve  y2  =  4  ax  at  the  origin. 

2.  Find  the  radius  of  curvature  of  the  curve  ys  +  xz  +  a(x2  +  y2) 
=  cfiy  at  the  origin. 

3.  Find  the  radius  of  curvature  of  the  curve  ahj  —  bxz  +  cx2y  at 
the  origin. 

Find  the  center  and  the  radius  of  curvature  for  each  of  the  following 
curves  at  the  point  (x,  y)  and  their  numerical  values  at  the  special 
point  indicated.  Find  where  the  curvature  is  greatest  and  least  on 
each  curve. 

4.  Rectangular  hyperbola  x$  =  m2  at  (m,  m). 

5.  Hyperbola  —  -  -^  =  1  at  (a,  0). 

a2     b2 


180 


DIFFERENTIAL   CALCULUS 


6.  General  parabola  an~xy  =  xn  at  (a,  a). 

7.  Parabola  Vx  +  Vy  =  Va  at  (a,  0). 

1  2  2 

8.  Hypocycloid  x3  -f  ys  =  of3"  at  the  point  at  which  x  =  y. 


9.   Cissoid  y" 


a,t  x  =  a. 


10.  Catenary  y—-(ea  _j_  e    <*)  at  £  =  0. 

11.  In  what  points  of  the  curve  y  =  x3  is  the  curvature  greatest  ? 


98.    Direct  derivation  of  the  expressions  for  k  and  R  in  polar 
coordinates.     Using  the  notation  of  Art.  58,  we  have 


hence 


But 


_d$_dB_ 

~  ds~  ds 

dO 


dO 
ds 
dO 


('+2) 


■'% 


tan^  =  p  — ,   ^=tan_1 
dp 


P_ 
dp 
dO 


(1) 


[Art.  44. 


therefore,  by  differentiating  as  to  0  and  reducing,  we  obtain 
'dp^2 


dij/  _  \d6 
d0~ 


P2  + 


(IT ' 


CONTACT   AND   CURVATURE  181 

which,  substituted  in  (1),  gives 


HUT 


Since  k  =  — ,  it  follows  that 
H 


R  = 


Mm 


fj 


*d02 


+  2 


(U 


This  result  should  be  compared  with  that  of  Art.  72. 

When  u  =  -   is   taken   as    dependent   variable,   the   expres- 
P 
sion  for  k  assumes  the  simpler  form 


K  =  


»r 


Since  at  a  point  of  inflexion  k  vanishes  and  changes  sign, 
hence  the  condition  for  a  point  of  inflexion,  expressed  in  polar 

72 

coordinates,  is  that  u  -\ shall  vanish  and  change  sign. 

cW 

EXERCISES 
Find  the  radius  of  curvature  for  each  of  the  following  curves : 

1.  p  =  a0.  3.   p  =  2  a  cos  $  -  a.       5.   p2  cos  2  0  =  a2. 

2.  p2  =  a2  cos  2  6.       4.   p  cos2  J  $  =  a.  6.  p  =  2  a(l  -  cos  0) . 

7.  P0  =  a. 


182  DIFFERENTIAL   CALCULUS 

EVOLUTES   AND   INVOLUTES 

99.  Definition  of  an  evolute.  When  the  point  P  moves  along 
the  given  curve,  the  center  of  curvature  C  describes  another 
curve  which  is  called  the  evolute  of  the  first. 

Let  f(x,  y)  =  0  be  the  equation  of  the  given  curve.  Then 
the  equation  of  the  locus  described  by  the  point  C  is  found  by 
eliminating  x  and  y  from  the  three  equations 


x—  a 


dy 
dx 


i+(& 

\dx 


dx2 


*-p- — it ' 

dx2 
and  thus  obtaining  a  relation  between  a,  /3,  the  coordinates  of 
the  center  of  curvature. 

No  general  process  of  elimination  can  be  given ;  the  method 
to  be  adopted  depends  upon  the  form  of  the  given  equation 

/(*,y)  =  o. 

Even  when  the  elimination  cannot  be  performed,  the  evolute 
can  be  traced  from  point  to  point  by  computing  successive 
values  of  (a,  /3)  corresponding  to  successive  values  of  (xf  y). 

Ex.  1.    Find  the  evolute  of  the  parabola  ?/2  =  4  px. 

Since  «=2pW,   ^  =  »H    <?t>  =  _  M*-! , 

dx  dx2  2 

hence        x  —  a  =  —  p^x~^(l  +  px~ *)2  p~?x%  —  —  2  (x  +  p), 
and  y  -  p  =  (1  +  px~x)2 p"~W  =  2(p~%x%  +  pW)  ; 

therefore  a  =  2p  +  3  a:,         ft  =  —  2  p*x*. 


CONTACT  AND   CURVATURE 


183 


Fig.  46 


The  result  of  eliminating  x  between  the  last  two  equations  is 

i.e.,  4(«-  2  Py  =  27  p/32, 

which  is  tho  equation  of  the  evolute  of  the  parabola,  a,  {3  being  the 

current  coordinates. 

Use  the  expressions  for  a  and  /3  to  compute  their  values,  and  to 
locate  the  points  («,  /3)  when  x  —  0, 


P 


Ex.  2.    Find  the  evolute  of  the  ellipse 

-2         ?<2 


Here 


dx2 


ir- 


x      ir 

a2      b2 


(1) 


y_  .dy  =  o    'hi-  - 


b2X 


b2     dx 


dx 


dy 


dx      -ft2/      ,  b2x2\      -b2,  2  „  ,   ,2  „.      -ft* 
r  a  #    \         a2y  /      a4#°  a2//3 


184  DIFFERENTIAL   CALCULUS 


whence 

y 


H  aW  \  b*        a>YJ      \  ¥  by- 


Therefore  -  (3  =  ^  ~  b' V-  (2) 

b* 

Similarly,  a  =  a—^x^.  (3) 

On   eliminating  x,  y  between   (1),  (2),  (3),  the  equation  of  the 
locus  described  by  (a,  (3)  is 

(aa)f  +  (bp)i  =  (a2  -  62)l  (Fig.  51) 

Use  (2),  (3)  to  locate  various  values  of  («,  /3),  and  trace  the  evolute. 

Take  a  =  2b;  a  =—. 

3 

100.    Properties  of   the  evolute.      The  evolute  has  two   im- 
portant properties  that  will  now  be  established. 

I.    The  normal  to  the  curve  is   tangent  to   the  evolute.     The 

relations  connecting  the  coordinates   (a,  ft)  of  the  center  of 

curvature  with   the   coordinates    (x,  y)  of   the  corresponding 
point  on  the  curve  are,  by  Art.  91, 

x-a  +  (y-/J)^  =  0,  (1) 

1+(!)'+(*-«g=°-  <2> 

By  differentiating  (1)  as  to  x,  considering  a,  (3,  y  as  functions 

of*,weobtain  +  ^_^_<L«_  M  =  0.      (3) 

\dxj       K       HJdx2     dx      dx  dx  w 

Subtracting  (3)  from  (2),  we  obtain 

da  +  dgdy  =  0^  ^ 

dx      dx  dx 

,  dB         dx 

whence  —  =  —  —  • 

da  dy 


CONTACT   AND   CURVATURE  185 

But  c-l=-  is  the  slope  of  the  tangent  to  the  evolute  at  (a,  /?), 
da 

and is  the  slope  of   the  normal  to  the  given  curve  at 

dy 

(x,  y).  Hence  these  lines  have  the 
same  slope ;  but  they  pass  through  the 
same  point  (a,  f3),  therefore  they  are 
coincident. 

II.    TJie  difference  between  two  radii     ' 
of  curvature  of  the  given   curve,  which 
touch  the   evolute  at  the  points   Clt  C2  Fig.  47 

(Fig.  47),  is  equal  to  the  arc  CXC2  of  the  evolute. 

Since  R  is  the  distance  between  the  points  (x,  y),  (a,  (3), 

hence  (aj  -  a)2  +  (y  -  /3)2  =  Br.  (5) 

When  the  point  (x,  y)  moves  along  the  given  curve,  the 
point  («,  /?)  moves  along  the  evolute,  and  thus  a,  /?,  R,  y,  are 
all  functions  of  x. 

Differentiation  of  (5)  as  to  x  gives 

(-^-SVMs-i?)-*^    (6) 


hence,  subtracting  (6)  from  (1),  we  obtain 

(X-a)p+(y-f})f  =  -lt™.  (7) 

dx  dx  dx 


Again,  from  (1)  and  (4), 


da  d(3 

dx  dx 

x  —  a     y  —  (3 


(8) 


186  DIFFERENTIAL   CALCULUS 

Hence,  each  of  these  fractions  is  equal  to 


4 


da\2,fdg\2  do- 

dx)       \dxj  dx 

V(a-a)2  +  (^=^_±^' 

in  which  <r  is  the  arc  of  the  evolute.     (Compare  Art.  41.) 

Next,  multiplying  numerator  and  denominator  of  the  first 
member  of  (8)  by  x  —  a,  and  those  of  the  second  member  by 
y  —  /?,  and  combining  new  numerators  and  denominators,  we 
find  that  each  of  the  fractions  in  (8)  is  equal  to 


(x-af  +  iy-py    ■ 

which  equals  — 

R0B 

dx 
R2 

by  (7)  and  (5). 

By  combining 

with 

(9),  we  obtain 

d<r_      dR 
dx          dx' 

that  is, 

^(cr±R)=0. 
dx 

Therefore  <r  ±  R  =  constant,  (10) 

wherein  o-  is  measured  from  a  fixed  point  A  on  the  evolute. 

Now,  let  (?!,  C2  be  the  centers  of  curvature  for  the  points 
Ply  P2  on  the  given  curve ;  let  PXCX  =  7^,  P2C2  =  R2 ;  and  let 
the  arcs  ACX,  AC2  be  denoted  by  p-lt  <r2.     Then 

<t1±R1  =  <t2±  R2,  by  (10) ; 

that  is  a-1  —  (T2=±  (R2  —  R}) ; 

hence,  arc  CXC2  —  R2  —  Rv  (11) 


CONTACT  AND   CURVATURE 


187 


Fig.  48 


Thus,  in  Fig.  48, 

P2C2+CA  =  PSC3,  etc. 

Hence,  if  a  thread  is  wrapped 
around  the  eve-lute,  and  then  is  un- 
wound, the  free  end  of  it  can  be  ^N 
made  to  trace  out  the  original  curve. 
From  this  property  the  locus  of  the 
centers  of  curvature  of  a  given 
curve  is  called  the  evolute  of  that  curve,  and  the  latter  is  called 
the  involute  of  the  former. 

When  the  string  is  unwound,  each  point  of  it  describes  a 
different  involute;  hence,  any  curve  has  an  infinite  number 
of  involutes,  but  only  one.  evolute. 

Any  two  of  these  involutes  intercept  a  constant  distance 
on  their  common  normal,  and  are  called  parallel  curves  on 
account  of  this  property. 

Ex.  Find  the  length  of  that  part  of  the  evolute  of  the  parab- 
ola which  lies  inside  the  curve. 

From  Fig.  46  the  required  length  is  twice  the  difference  between 
the  tangents  C3P3  and  PQC0,  both  of  which  are  normals  to  the 
parabola. 

To  find  the  coordinates  of  the  point  P3,  write  the  equation  of  the 
tangent  to  the  evolute  at  C3,  and  find  the  other  point  at  which  it 
intersects  the  parabola. 

The  coordinates  of  C3,  the  point  of  intersection  of  the  two  curves, 
are  (8p,  4joV2),  and  the  equation  of  the  tangent  at  C3  is 

2x  -V2y  -  Sp  =  0. 

This  tangent  intersects  the  parabola  at  the  point  (2p,  —  2  V2 />), 
which  is  Po. 


188  DIFFERENTIAL   CALCULUS 

The  value  of  the  radius  of  curvature  is  ~{x  ~*~P)   f  hence  P0C0=2p, 

Vp 
P3C3  =  GV3p,  hence  the  arc  C0CS  is  2p(3V3  -  1),  and  the  required 

length  of  the  evolute  is  therefore  4jt>(3V3  —  1). 

EXERCISES 

Find  the  coordinates  of  the  center  of  curvature  for  each  of  the 
following  curves : 

1.    x2  +  y2  =  a2.  3.    y3  ^  a2x. 


2.    x  =  a  log  a  +  ^a'2  ~y2  -  Va2  -  y2.      4.   y=-^(ea  +  e   «). 


Find  the  equations  of  the  evolutes  of  the  following  curves : 

5.    xy  =  a2.  6.    a2y2  -  b2x2  =  -  a2b2.  7.    x%  +  y*  =  ah 

8.  Show  that  the  curvature  of  an  ellipse  is  a  minimum  at  the  end 
of  the  minor  axis,  and  that  the  osculating  circle  at  this  point  has  con- 
tact of  the  third  order  with  the  curve. 


Fia.  49 


This  circle  of  curvature  must  be  entirely  outside  the  ellipse 
(Fig.  49).  For,  consider  two  points  Pi,  P2,  one  on  each  side  of  B, 
the  end  of  the  minor  axis.     At  these  points  the  curvature  is  greater 


CONTACT   AND   CURVATURE 


189 


than  at  B,  hence  these  points  must  be  farther  from  the  tangent  at  B 
than  the  circle  of  curvature,  which  has  everywhere  the  same  curva- 
ture as  at  B. 

9.  Similarly,  show  that  the  curvature  at  A,  the  end  of  the  major 
axis,  is  a  maximum,  and  that  the  circle  of  curvature  at  A  lies  entirely 
within  the  ellipse  (Fig.  49). 

10.  Show  how  to  sketch  the  circle  of  curvature  for  points  between 
A  and  B.  The  circle  of  curvature  for  points  between  A  and  B  has 
three  coincident  points  in  common  with  the  ellipse  (Art.  93),  hence 
the  circle  crosses  the  curve  (Art.  89).  Let  K,  P,  L  be  three  points 
on  the  arc,  such  that  K  is  nearest  A  and  L  nearest  B.     The  center 


Fig.  50 

of  curvature  for  P  lies  on  the  normal  to  P,  and  on  the  concave  side 
of  the  curve.  The  circle  crosses  at  P,  lying  outside  of  the  ellipse  at 
K  (on  the  side  towards  .4),  and  inside  the  ellipse  at  L;  for  the  bend- 
ing of  the  ellipse  increases  from  B  to  P  and  from  P  to  K,  while  the 
bending  (curvature)  of  the  osculating  circle  remains  constant 
(Fig.  50). 

11.   Two  centers  of  curvature  lie  on  every  normal.     Prove  geo- 
metrically that  the  normals  to  the  curve  are  tangent  to  the  e volute. 


190  DIFFERENTIAL   CALCULUS 

12.    Show  that  the  entire  length  of  the  e volute  of  the  ellipse  is 

4  (- —  —  ).     [From  equation  (11)  above,  take  Bv  R2  as  the  radii  of 

\b       a  J 

curvature  at  the  extremities  of  the  major  and  minor  axes.] 

13.  If  E  is  the  center 
of  curvature  at  the  vertex 
A  (Fig.  51),  prove  that 
CE  =  ae1,  in  which  e  is 
the  eccentricity  of  the 
ellipse;  and  hence  that 
CD,  CA,  CF,  CE  form 
a  geometric  series  whose 
common  ratio  is  e.  Show 
also  that  DA,  AF,  FE 
form  a  similar  series. 

14.  If  H  is  the  center 
of  curvature  for  B,  show 
that  the  point  77  is  with- 
out or  within  the  ellipse, 
according       as       a  >  or 

<  bV2,  or  according  as  e2  >  or  <  J.     Sketch  the  evolute  when  b  =  -— -• 

o 

15.  Show  by  inspection  of  the  figure  that  four  real  normals  can 
be  drawn  to  the  ellipse  from  any  point  within  the  evolute. 

16.  Find  the  parametric  equations  of  the  evolute  of  the  cycloid 

x  =  a(d  —  sin  $),   y  =  a  (I  —  cos  6). 


CHAPTER   XIII 

SINGULAR   POINTS 

101.  Definition  of  a  singular  point.    If  the  equation  f(x,  y)  =  0 

is  represented  by  a  curve,  the  derivative  -^,  when  it  has  a 

cix 

determinate  value,  measures  the  slope  of  the  tangent  at  the 
point  (x,  y).  There  may  be  certain  points  on  the  curve,  how- 
ever, at  which  the  expression  for  the  derivative  assumes  an 
illusory  or  indeterminate  form ;  and,  in  consequence,  the  slope 
of  the  tangent  at  such  a  point  cannot  be  directly  determined 
by  the  method  of  Art.  5.  Such  values  of  x,  y  are  called  sin- 
gular values,  and  the  corresponding  points  on  the  curve  are 
called  singular  points. 

102.  Determination  of  singular  points  of  algebraic  curves. 
When  the  equation  of  the  curve  is  rationalized  and  cleared  of 
fractions,  let  it  take  the  form /(a?,  y  =  0. 

This  gives,  by  differentiation  with  regard  to  x,  as  in  Art.  65, 

df  +  dfdy^Q 
dx      dy  dx 

3f 
du  dr 

whence  Tx  =  ~W'  (1) 

By 

In  order  that  —  may  become  illusory,  it  is  therefore  neces- 
dx 

sarytohave  #  =  0,    ^  =  0.  (2) 

dx  dy 

191 


192  DIFFERENTIAL   CALCULUS 

Thus,  to  determine  whether  a  given  curve  f(x,  y)  =  0  has 

singular  points,  put  J-  and  f-  each  equal  to  zero  and  solve 
dx  dy 

these  equations  for  x  and  y. 

If  any  pair  of  values  of  x  and  y,  so  found,  satisfy  the  equa- 
tion f(x,  y)  =  0,  the  point  determined  by  them  is  a  singular 
point  on  the  curve. 

To  determine  the  appearance  of  the  curve  in  the  vicinity 
of  a  singular  point  (xlf  yx),  evaluate  the  indeterminate  form 

dy  _      dx      0 
dx  =  ~aj=^ 

by  finding  the  limit  approached  continuously  by  the  slope  of 
the  tangent  when  x  =  xx,  y  =  yv  "** 

dy  dx\dxy 


Hence 


dx  d  fdf 

dx 


dx2      dx  dydx 

d2f    ,  d*fdy 

dx  dy      dy2dx 


[Arts.  64,  85. 


at  the  point  (xly  y{). 

This  equation  cleared  of  fractions  gives,  to  determine  the 
slope  at  (ajj,  ?/i)>  the  quadratic 

dy2[dxj         dxdy\dx)      dx2  V  } 

This    quadratic    equation    has   in   general   two   roots.     The 
only  exceptions  occur  when  simultaneously,  at  the   point  in 


SINGULAR   POINTS  193 

in  which  case  -^  is  still  indeterminate  in  form,  and  must  be 
dx 

evaluated  as  before.     The  result  of  the  next  evaluation  is  a 

cubic  in  — ,  which  gives  three  values  to  the  slope,  unless  all 
dx 

the  third  partial  derivatives  vanish  simultaneously  at  the 
singular  point. 

The  geometric  interpretation  of  the  two  roots  of  equation 
(3)  will  now  be  given,  and  similar  principles  will  apply  when 
the  quadratic  is  replaced  by  an  equation  of  higher  degree. 

The  two  roots  of  (3)  are  real  and  distinct,  real  and  coinci- 
dent, or  imaginary,  according  as 

Hf  a2/y     d2f  d*f 

\dx  dyj       dx2  By2 

is  positive,  zero,  or  negative.  These  three  cases  will  be  con- 
sidered separately. 

103.    Multiple  points.     First  let  77  be  positive.     Then  at  the 

point  (x,  y)  for  which  -J-  =  0,   -J-  =  0,  there  are  two  values  of 
dx  dy 

the  slope,  and  hence  two  distinct  singular  tangents.  It  fol- 
lows from  this  that  the  curve  goes  through  the  point  in  two 
directions,  or,  in  other  words,  two  branches  of  the  curve  cross 
at  this  point.  Such  a  point  is  called  a  real  double  point  of 
the  curve,  or  simply  a  node.  The  conditions,  then,  to  be  satis- 
fied at  a  node  (xlf  y{)  are 

oxv  oyL 

and  H(xu  yx)  >  0. 

Ex.     Examine  for  singular  points  the  curve 

3  x2  _  xy  _  o  f  +  xs  -8y*  =  0. 

EL.    CALC.  —  13 


194  DIFFERENTIAL   CALCULUS 

Here  $f=6x-y+3x*,    &  =  -  x  -  4  y  -  21  v*. 

dx  dy  9  9 

The  values  x  =  0,  y  =  0  will  satisfy  these  three  equations,  hence 
(0,  0)  is  a  singular  point. 


Since 


|^=6+  Qx  =  6  at  (0,0), 


Fig.  52 

hence  the  equation  determining  the  slope  is,  from  (3), 

-*a)"-«e)+-* 

of  which  the  roots  are  1  and  —  f .     It  follows  that  (0,  0)  is  a  double 
point  at  which  the  tangents  have  the  slopes  1,  —  f. 

Find  the  equation  of  the  real  asymptote,  and  the  coordinates  of  the 
finite  point  in  which  it  meets  the  curve. 

104.  Cusps.  Next  let  H=0.  The  two  tangents  are  then 
coincident,  and  there  are  two  cases  to  consider.  If  the  curve 
recedes  from  the  tangent  in  both  directions  from  the  point  of 
tangency,  the  singular  point  is  called  a  tacnode.  Two  distinct 
branches  of  the  curve  touch  each  other  at  this  point.  (See 
Fig.  53.) 


SINGULAR  POINTS 


195 


If  both  branches  of  the  curve  recede  from  the  tangent  in 
only  one  direction  from  the  point  of  tangency,  the  point  is 
called  a  cusp. 

Here  again  there  are  two  cases  to  be  distinguished.     If  the 
branches  recede  from  the  point  on  opposite  sides  of  the  double 
tangent,  the  cusp  is  said  to  be  of  the  first  kind ;  if  they  recede  ( 
on  the  same  side,  it  is  called  a  cusp  of  the  second  kind. 

The  method  of  investigation  will  be  illustrated  by  a  few 
examples. 

.     Ex.  1.  f(x,  y)  =  aY  -  <*2xA  +  *6  =  0. 

^  =  -4a2z«  +  6a?s;  &  =  2  a*y. 
dx  By 

The  point  (0,  0)  will  satisfy /(x,  y)  =  0,  $£■  =  0,  |^=  0;  hence  it 

dx  dy 

is  a  singular  point.     Proceeding  to  the  second  derivatives,  we  obtain 


§y  =  -  12  a2*2  +  30  x*  =  0  at  (0,  0), 

dx2 


BY  = 
dxd/ 

a2/= 


o, 


The  two  values  of  -*  are  there- 
fore coincident,  and  each  equal  to  Fig.  53 
zero.     From  the  form  of  the  equation,  the  curve  is  evidently  sym- 
metrical with  regard  to  both  axes ;  hence  the  point  (0,  0)  is  a  tacnode. 

No  part  of  the  curve  can  be  at  a  greater  distance  from  the  y-axis 
than  ±  a,  at  which  points  -^  is  infinite.     The  maximum  value  of  y 

corresponds  tox  =  ±  aVf.     Between  x  =  0,  x  =  aVj  there  is  a  point 
of  inflexion  (Fig.  53). 


196  DIFFERENTIAL   CALCULUS 

Sketch  the  curves  obtained  by  giving  larger  and  larger  values  to  the 
parameter  a. 

Ex.2.    f(x,y)  =  y*-x*  =  0; 

dx~      6X'dy-2y' 
Hence  the  point  (0,  0)  is   a  singular  point. 

Further,  ^  =  -  6a:  =  0  at   (0,  0)  ; 

d2f  =0.  dY=2 
dxdy       '  dy2 

Therefore  the  two  roots  of  the  quadratic  equation  defining  -^  are 

both  equal  to  zero.  So  far,  this  case  is  exactly  like  the  last  one,  but 
here  no  part  of  the  curve  lies  to  the  left  of  the  axis  y.  On  the  right 
side,  the  curve  is  symmetric  with  regard  to  the  x-axis.  As  x  increases, 
y  increases  ;  there  are  no  maxima  nor  minima,  and  no  inflexions 
(Fig.  54).       . 

p]x.  3.  f(x,  y)  -  x4  -  2  ax'2y  -  axy2  -f  ahf-  =  0. 

The  point  (0,  0)  is  a  singular  point,  and  the  roots  of  the  quadratic 
defining  -j-  are  both  equal  to  zero,  hence  the  origin  is  a  cusp,  and  the 

cuspidal  tangent  is  the  a; -axis. 

To  show  the  form  of  the  curve  near  the  cusp,  solve  the  equation 
for  ?/.     Then  ._ 

a  —  x\         ya  I 
First  suppose  that  a  is  positive. 

When  x  is  negative,  y  is  imaginary ;  when  x  =  0,  y  =  0 ;  when  x  is 
positive,  but  less  than  a,  y  has  two  positive  values,  therefore  two 
branches  are  above  the  a>axis.  When  x  =  a,  one  branch  becomes  in- 
finite, having  the  asymptote  x  =  a ;  the  other  branch  has  the  ordinate 
\  a.     The  origin  is  therefore  a  cusp  of  the  second  kind  (Fig.  55). 

Next  suppose  that  a  is  negative.  When  x  is  positive,  y  is  imagi- 
rary;  when  x  is  negative,  y  is  real.     The  same  reasoning  as  before 


SINGULAR   POINTS 


197 


shows  that  there  is  a  cusp  of  the  second  kind  in  the  second  quarter, 
with  the  #-axis  as  a  cuspidal  tangent. 

Examine  the  transition  case  in  which  a  =  0. 


Fig.  54 


Fig.  55 


105.  Conjugate  points.  Lastly,  let  H  be  negative.  In  this 
case  there  are  no  real  tangents  ;  hence  no  points  in  the  im- 
mediate vicinity  of  the  given  point  satisfy  the  equation  of  the 
curve. 

Such  an  isolated  point  is  called  a  conjugate  point. 


Ex.  1.  /(*,  y)  =  ay2  -  x*  +  bx2  =  0. 

Here  (0,  0)  is  a  singular  point  of  the 
locus,  and  at  this  point  we  find 

dy  = 

dx 

both  roots  being  imaginary  if  a  and  b 
have  the  same  sign. 

To  show  the  form  of  the  curve,  solve 
the  given  equation  for  y. 


Then 


y=±x\ 


x  —  b 


Fig.  5G 


and  hence,  if  a  and  b  are  positive,  there  are  no  real  points  on  the 
curve  between  x  =  0  and  x  -  b.    Thus  O  is  an  isolated  point  (Fig.  56). 
Examine  the  cases  in  which  a  or  b  is  negative. 


198  DIFFERENTIAL   CALCULUS 

These  are  the  only  singular  ities  that  algebraic  curves  can 
have,  although  complicated  combinations  of  them  may  appear. 
In  each  of  the  foregoing  examples,  the  singular  point  was 
(0,  0) :  but  for  any  other  point,  the  same  reasoning  will  apply. 

Ex.  2.      f(x,  y)  =  x2  +  3  y*  -  13  y2  -  4  x  +  17  y  -  3  =  0, 

3£=2.r-4,    3/"=  9  y2- 26  y +  17. 
dx  dy 

At  the  point  (2,  1),  /(2,  1)  =  0,    &=0,    ^=0;   hence  (2,  1)  is 
,  .  4.  dx  dy 

a  singular  point. 

Also^  =  2;    JE£-=0;    |2f=  18y  -  26,   =  -  8  at  (2,  1). 

dx2  da;  cty  d//2 

Hence  —  =  ±  -  ;  and  thus  the  equations  of  the  two  tangents  at  the 
dx  2 

node  (2, 1)  are  y  -  1  =  \(x  -  2),  y  -  1  =  -  |  (a:  -  2). 

When  JT  is  negative,  the  singular  point  is  necessarily  a  con- 
jugate point,  but  the  converse  is  not  always  true.  A  singular 
point  may  be  a  conjugate  point  when  H  =  0.  [Compare 
Ex.  4  below.] 

EXERCISES  ON   CHAPTER  XIII 

Examine  each  of  the  following  curves  for  multiple  points  and  find 
the  equations  of  the  tangents  at  each  such  point ;  also  find  the 
asymptotes  and  sketch  the  curve  : 

1.    a2x2  =  b2y2  +  *V- 


2  a  —  x 

3.  x3  +  y\  =  «f ;  or,  in  rational  form,  (x2+y2  —  a2)3 +  27  a2x2y2  =  0. 

4.  y2(x2  -  a2)  =  x4. 

5.  y—  a+  x  +  bx2  +  cx%  ;  or,  in  rational  form, 

( y  -a-x-  bx2)  2  -  c2s«  =  ( ) . 


SINGULAR   POINTS  199 

When  a  curve  has  two  parallel  asymptotes  it  is  said  to  have  a  node 
at  infinity  in  the  direction  of  the  parallel  asymptotes.  Apply  to 
Ex.  6. 

6.  (x*-y*y—4y*  +  y  =  0. 

7.  x*-2  ay*  -  3  a2y2  -  2  d2x2  +  a4  =  0. 
J.   y2  _  x(x  +  a)2.  a  >  o .  a  <  o. 

9.   xs  —  3  axy  +  yz  =  0.     Find  the  asymptote  and  sketch  the  curve. 

10.  y'2  =  x*  +  x5. 

11.  Show  that  the  curve  y  =  x  log  a:  has  a  terminating  point  at  the 
origin.     Find  the  minimum  value  of  y  and  sketch  the  curve. 

12.  y  —  x2  log  x. 


CHAPTER   XIV 

ENVELOPES 
106.    Family  of  curves.     The  equation  of  a  curve, 

/(»,y)  =  o, 

usually  involves,  besides  the  variables  x  and  ?/,  certain  coeffi- 
cients that  serve  to  fix  the  size,  shape,  and  position  of  the 
curve.  The  coefficients  are  called  constants  with  reference 
to  the  variables  x  and  y,  but  it  has  been  seen  in  previous 
chapters  that  they  may  take  different  values  in  different 
problems,  while  the  form  of  the  equation  is  preserved.  Let 
a  be  one  of  these  "constants."  Then  if  a  is  given  a  series 
of  numerical  values,  and  if  the  locus  of  the  equation,  corre- 
sponding to  each  special  value  of  a  is  traced,  a  series  of  curves 
is  obtained,  all  having  the  same  general  character,  but  differ- 
ing somewhat  from  each  other  in  size,  shape,  or  position.  A 
system  of  curves  so  obtained  is  called  a  family  of  curves. 

For  example,  if  h,  k  are  fixed,  and  }>  is  arbitrary,  the  equa- 
tion (y—k)2  =  2p(x  —  h)  represents  a  family  of  parabolas, 
each  curve  of  which  has  the  same  vertex  (h,  k),  and  the  same 
axis  y  =  k,  but  a  different  latus  rectum.  Again,  if  k  is  the 
arbitrary  constant,  this  equation  represents  a  family  of  parab- 
olas having  parallel  axes,  the  same  latus  rectum,  and  having 
their  vertices  on  the  same  line  x  =  h. 

The  presence  of  an  arbitrary  constant  a  in  the  equation  of 
a  curve  is    indicated    in    functional    notation    by  writing   the 

200 


ENVELOPES  201 

equation  in  the  form ,/(#,  y,  a)  =  0.  The  quantity  a,  which 
is  constant  for  the  same  curve  but  different  for  different 
curves,  is  called  the  parameter  of  the  family.  The  equations 
of  two  neighboring  curves  are  then  written 

f(x,  y,  a)  =  0,  />,  y,  a  +  h)  =  0, 

in  which  h  is  a  small  increment  of  a.  These  curves  can  be 
brought  as  near  to  coincidence  as  desired  by  diminishing  h. 

107.  Envelope  of  a  family  of  curves.  A  point  of  intersection 
of  two  neighboring  curves  of  the  family  tends  toward  a  limit- 
ing position  as  the  curves  approach  coincidence.  The  locus  of 
such  limiting  points  of  intersection  is  called  the  envelope  of 
the  family. 

Let  f(x,  y,  a)  =  0,  f(x,  y,  a  +  h)  =  0  (1) 

be  two  curves  of  the  family.  By  the  theorem  of  mean  value 
'(Art.  39) 

f(x,  y,  a  +  h)  =  f(x,  y,  a)  +  h d/-  (as,  y,  a  +  6k),  (2) 

da 

which,  on  account  of  equation  (1),  reduces  to 
d£{x,y,«  +  6h)=(). 

Hence,  it  follows  that  in  the  limit,  when  h  =  0, 

J-(x,  y,  «)=0 
da 

is  the  equation  of  a  curve  passing  through  the  limiting  points 
of  intersection  of  the  curve /(a?,  y}  a)  =  0  with  its  consecutive 
curve.  This  determines  for  any  assigned  value  of  a  definite 
limiting  points  of  intersection  on  the  corresponding  member  of 


202  DIFFERENTIAL  CALCULUS 

the  family.    The  locus  of  all  such  points  is  then  to  be  obtained 
by  eliminating  the  parameter  a  from  the  equations 

f(x,  y,  a)  =  0,    -f  (x,  y,  a)  =  0. 
da 

The  resulting  equation  in  x  and  y  represents  the  fixed  enve- 
lope of  the  family. 

108.   The  envelope  touches  every  curve  of  the  family. 

I.  Geometrical  proof.  Let  A,  B,  C  (Fig.  57)  be  three  consec- 
utive curves  of  the  family ;  let  A,  B  intersect  in  P,  and  B,  C  inter- 
sect in  Q.  When  P,  Q  approach  coincidence,  PQ  will  be  the 
direction  of  the  tangent  to  the  envelope  at  P;  but  since  P,  Q 


are  two  points  on  B  that  approach  coincidence,  hence  PQ  is 

also  the  direction  of  the  tangent  to  B  at  P,  and  accordingly  B 

and  the  envelope  have  a  common  tangent  at  P.     Similarly  for 

every  curve  of  the  family. 

II.     More  rigorous  analytical  proof.     Let   —  f(x,  ?/,  a)  =  0 

da 

be  solved  for  a,  in  the  form  a  =  <£(#,  y).     Then  the  equation 

of  the  envelope  is 

f(x,  y,  <f>(x,  y))  =  0. 

Equating  the  total  avderivative  to  zero,  we  obtain 
dx      dy  dx      d<f>\dx      By  dxj 


ENVELOPES  203 

but  -J-  =  -J-  =  0,   hence  the  slope  of  the  tangent  to  the  enve- 
d<f>      da 

lope  at  the  point  (x,  y)  is  given  by 

df+dfdy  =  0 
dx      dy  dx 

But  this  equation  defines  the  direction  of  the  tangent  to  the 
curve /(a?,  y,  a)  —  0  at  the  same  point,  and  therefore  a  limit- 
ing point  of  intersection  on  any  member  of  the  family  is  a 
point  of  contact  of  this  curve  with  the  envelope. 

Ex.  Find  the  envelope  of  the  family  of  lines 


Differentiate  (1)  as  to  m, 


y  =  mx+£-,  (1) 

•  m 

obtained  by  varying  m. 


0  =*-■£.  (2) 

To  eliminate  m  multiply  (2)  by  m  and  square ;  square  (1)  and  sub- 
tract the  first  from   the  second.      The  envelope  is  found  to  be  the 

parabola  «     A 

r  y2=  Ipx. 

Draw  the  lines  (1)  corresponding  to 

m  =  1,  2,  3,  4,  oo  ;  m  -  -  1,  -  2,  -  3,  -  4. 

109.  Envelope  of  normals  of  a  given  curve.  The  evolute 
(Art.  99)  was  defined  as  the  locus  of  the  centers  of  curvature. 
The  center  of  curvature  was  shown  to  be  the  point  of  intersec- 
tion of  consecutive  normals  (Art.  92),  whence  by  Art.  107  the 
envelope  of  the  normals  is  the  evolute. 

Ex.   Find  the  envelope  of  the  normals  to  the  parabola  y2  =  ±px. 
The  equation  of  the  normal  at  (xv  y\)  is 


204  DIFFERENTIAL   CALCULUS 

or,  eliminating  x\  by  means  of  the  equation  yf  =  ipxv  we  obtain 

J      Ul      8^     2p  K  } 

The  envelope  of  this  line,  when  yx  takes  all  values,  is  required. 

Differentiate  as  to  ?/,,  n     „ 

_  i  -  ^JL\  _  JL 
8p*      2p' 

On  substituting  this  value  for  yl  in  (1),  the  result, 

27 pif  =  4(>  -  2  jo)3, 

is  the  equation  of  the  required  evolute.     Show  that  this  semi-cubical 
parabola  has  a  cusp  at  (2p,  0).     Trace  the  curve. 

110.  Two  parameters,  one  equation  of  condition.  In  many 
cases  a  family  of  curves  may  have  two  parameters  which  are 
connected  by  an  equation.  For  instance,  the  equation  of  the 
normal  to  a  given  curve  contains  two  parameters  xYi  yx  which 
are  connected  by  the  equation  of  the  curve.  In  such  cases 
one  parameter  may  be  eliminated  by  means  of  the  given  rela- 
tion, and  the  other  treated  as  before. 

When  the  elimination  is  difficult  to  perform,  both  equations 
may  be  differentiated  as  to'  one  of  the  parameters,  a,  regard- 
ing the  other  parameter  /?  as  a  function  of  a.     This  gives  four 

equations   from  which  a,  8,  and  —   may  be  eliminated,  the 

da 

resulting  equation  being  that  of  the  desired  envelope. 
Ex.  1.     Find  the  envelope  of  the  line 

a      b 
the  sum  of  its  intercepts  remaining  constant. 


ENVELOPES    ■  205 


The  two  equations  are         -  -\-  -J-  —  1, 
a      b 

a  +  b  =  c. 

Differentiate  both  equations  as  to  a ; 

—  x  _y_db  _  „ 
•      a2       H2da~    ' 


1+* 

da 


Eliminate 


Then  —  =  *-.  which  reduces  to 
a2      b2 


dbi 
da 


x      y      x  +  y 
a      b      a      h 


-  —  - ;  whence  a  =  Vex,  b  =  Vey. 


a      b      a  +  b      c 

Therefore  Vx  +  Vy  =  Vc 

is  the  equation  of  the  desired  envelope.     [Compare  Ex.  p.  87.] 
This  equation  when  rationalized  is 

(x-  yy-2c(x  +  y)  +  c2  =  0. 

By  turning  the  coordinate  axes  through  45°,  show  that  this  repre- 
sents a  parabola  whose  axis  bisects  the  angle  between  the  original 
axes.  Show  that  the  curve  touches  both  these  axes.  Draw  different 
lines  of  the  family,  corresponding  to  a  =  4,  b  =  4 ;  a  =  5,  b  =  3 ;  a  =  6, 
Z>  =  2 ;  a  -  7,  b  =  1 ;  a  =  8,  6  =  0;  etc. 

Ex.  2.     Find  the  envelope  of  the  family  of  coaxial  ellipses  having 

a  constant  area. 

Here  x-  +  f-  =  \- 

a* l      b2 

a&  =  £2. 

For  symmetry,  regard  a  and  6  as  functions  of  a  single  parameter  /. 


206  DIFFERENTIAL   CALCULUS 

Then  by  differentiation  as  to  t, 


&<]±jrt_db  _0 
as  dt      b*  dt.~    ' 


hence 


-.da  .      db      A 

b h  a —  =  0 

dt         dt 


&  =  T 


i2      62     2 


a  =  ±xV'2,   b  =±yV2, 
and  the  envelope  is  the  pair  of  rectangular  hyperbolas  xy  =  ±  \  k'2. 

Y 


Fig.  58 


Note.  A  family  of  curves  may  have  no  envelope;  i.e.,  consecutive 
curves  may  not  intersect;  e.g.,  the  family  of  concentric  circles  x2  +y2 
=  r2,  obtained  by  giving  r  all  possible  values. 


ENVELOPES  207 

If  every  curve  of  a  family  has  a  node,  and  the  node  has 
different  positions  for  different  curves  of  the  family,  the  enve- 
lope will  be  composed  of  two  (or  more)  curves,  one  of  which 
is  the  locus  of  the  node. 

Ex.   Find  the  envelope  of  the  system 

in  which  X  is  a  varying  parameter. 

Here  3L  =  —  2(y  —  A.)  =  0;  by  combining  with  /=  0  to  eliminate 
dX 

X,  we  obtain  x2  =  0,   x  —  1  =  0,   x  +  1  =  0. 

From  Art.  103  it  is  seen  that  the  point 
x  =  0,  y  =  X 
is  a  node  on  /;  moreover,  the  various  curves  of  the  family  are  ob- 
tained by  moving  any  one  of  them  parallel  to  the  y-axis.     The  lines 
x  —  1=0,  £  +  1  =  0  form  the  proper  envelope,  and  x  =  0  is  the  locus 
of  the  node. 

EXERCISES  ON   CHAPTER  XIV 

Find  the  envelope  of  each  of  the  following  families  of  curves ; 
draw  to  scale  various  members  of  the  family,  and  verify  that  the  en- 
velope has  been  correctly  found. 

1.  The  family  of  straight  lines  x  cos  a  +  y  sin  a  =  p,  when  «  is  a 
parameter. 

2.  A  straight  line  of  fixed  length  a  moving  with  its  extremities 
in  two  rectangular  axes. 

3.  Ellipses  described  with  common  centers  and  axes,  and  having 
the  sum  of  the  semi-axes  equal  to  c. 

4.  The  straight  lines  having  the  product  of  their  intercepts  on 
the  coordinate  axes  equal  to  k2. 

5.  The  lines  y  -  ft  =  m  (x  -«)+  rvT+m*,  m  being  a  variable 
parameter. 


208  DIFFERENTIAL   CALCULUS 

6.  A  circle   moving  with  its  center  on  a  parabola  whose  equation 
is  y2  =  4  ax,  and  passing  through  the  vertex  of  the  parabola. 

7.  A  perpendicular   to    any    normal  to    the   parabola   y2  =  4  ax, 
drawn  through  the  intersection  of  the  normal  with  the  x-axis. 

8.  The  family  of  circles  whose  diameters  are  double  ordinates  of 
the  ellipse  b2x'2  +  a2y2  =  a2//2. 

9.  The  circles  which   pass  through   the   origin    and   have   their 
centers  on  the  hyperbola  x2  —  y2  =  c2. 

10.  The  family  of  straight  lines  y  =  2  mx  +  mA,  m  being  the  vari- 
able parameter. 

11.  The  ellipses  whose  axes  coincide,  and  such  that  the  distance 
between  the  extremities  of  the  major  and  minor  axes  is  constant  and 
equal  to  k. 

12.  From  a  fixed  point  on  the  circumference  of  a  circle  chords  are 
drawn,  and  on  these  as  diameters  circles  are  described. 

13.  With  the  point  (xi,  //i)  on  a  given  ellipse  as  center,  an  ellipse 
is  described  having  its  axes  equal  and  parallel  to  those  of  the  given 
ellipse.     Let  (xi,  y{)  describe  the  given  ellipse. 

14.  Show  that  if  the  corner  of  a  rectangular  piece  of  paper  is 
folded  down  so  that  the  sum  of  the  edges  left  unfolded  is  constant, 
the  crease  will  envelop  a  parabola. 

15.  In  the  "  nodal  family  "  (y  -  2  a)2=(x  -  «)2+  8  xs  -  y3,  show 
that  the  usual  process  gives  for  envelope  a  composite  locus,  made  up 
of  the  "node-locus"  (a  line)  and  the  envelope  proper  (an  ellipse). 

16.  The  family  of  curves  (y  —  x2)  +  a  (x  —  y2)  =  0. 


INTEGRAL   CALCULUS 


:>>*< 


CHAPTER   I 

GENERAL  PRINCIPLES  OF  INTEGRATION 

111.  The  fundamental  problem.  The  fundamental  problem 
of  the  Differential  Calculus,  as  explained  in  the  preceding 
pages,  is  this : 

Given  a  function  f(x)  of  an  independent  variable  x,  to  deter- 
mine its  derivative  f'(x). 

It  is  now  proposed  to  consider  the  inverse  problem,  viz. : 
Given  any  function  f'(x),  to  determine  the  function  f(x)  hav- 
ing f'(x)  for  its  derivative. 

The  solution  of  this  inverse  problem  is  one  of  the  objects 
of  the  Integral  Calculus. 

The  given  function  f(x)  is  called  the  integrand,  the  func- 
tion f(x)  which  is  to  be  found  is  called  the  integral,  and  the 
process  gone  through  in  order  to  obtain  the  unknown  function 
f{x)  is  called  integration. 

The  operation  and  result  of  differentiation  are  symbolized 
by  the  formula  cj 

Txf{x)=m'  (1) 

or,  written  in  the  notation  of  differentials, 

df(x)=f\x)dx.  (2) 

el.  c/xc.  — 14  209 


210  INTEGRAL   CALCULUS 

The  operation  of  integration  is  indicated  by  prefixing  the 
symbol    j   to  the  function,  or  differential,  whose  integral  it  is 

required  to  find.  It  is  called  the  integral  sign,  or  the  sign  of 
integration.  Accordingly,  the  formula  of  integration  is  written 
thus : 


/(*)=//'(*) 


dx. 


Following  long  established  usage,  the  differential,  rather 
than  the  derivative,  of  the  unknown  function  /(a?)  is  written 
under  the  sign  of  integration.  One  of  the  advantages  of  so 
doing  is  that  the  variable,  with  respect  to  which  the  integration 
is  performed,  is  explicitly  mentioned.  This  is,  of  course,  not 
necessary  when  only  one  variable  is  involved,  but  it  is  essential 
when  several  variables  enter  into  the  integrand,  or  when  a 
change  of  variable  is  made  during  the  process  of  integration. 

112.  Integration  by  inspection.  The  most  obvious  aid  to 
integration  is  a  knowledge  of  the  rules  and  results  of  differen- 
tiation. It  frequently  happens  that  the  required  function /(a?) 
can  be  determined  at  once  by  recollecting  the  result  of  some 
previous  differentiation. 

For  example,  suppose  it  is  required  to  find 


/ 


cos  x  dx. 


It  will  be  recalled  that  cos  x  dx  is  the  differential  of  sin  x,  and 
thus  the  proposed  integration  is  immediately  effected ;  that  is, 


/ 


cos  xdx  =  sin  x. 


Again,  suppose  it  is  required  to  integrate 

xndx, 


/• 


GENERAL   PRINCIPLES   OF   INTEGRATION  211 

in  which  n  is  any  constant  (except  —  1).  This  problem  sug- 
gests the  formula  for  differentiating  a  variable  affected  by  a 
constant  exponent  [(6),  p.  44].     The  formula  referred  to  may 

be  written  .    n+1 

d(- )=  xndx, 

\n  +  y 

/»  zy.n  +  1 

and  hence  we  conclude, 


s,     f  xn  dx 


?i  +  l 


An  exception  to  this  result  occurs  when  n  has  the  value  —  1. 

For  in  that  case  we  deduce  from  (8),  p.  44,  the  formula  of 

integration 

I  x  L  dx  =  I  —  =  log  x. 

The  method  used  in  the  above  illustrations  may  be  designated 
as  integration  by  inspection.  This  is,  in  fact,  the  only  practical 
method  available.  The  object  of  the  various  devices  suggested 
in  the  subsequent  pages  is  to  transform  the  given  integrand 
or  to  separate  it  into  simpler  elements  in  such  a  way  that  the 
method  of  inspection  can  Ipe  applied. 

113.  The  fundamental  formulas  of  integration.  When  the 
formulas  of  differentiation,  pp.  44-45,  are  borne  in  mind,  the 
method  of  inspection  referred  to  in  the  preceding  article  leads 
at  once  to  the  following  fundamental  integrals.  Upon  these, 
sooner  or  later,  every  integration  must  be  made  to  depend. 


clu  = 


n  +  l 


n+  1 


I.     f  utl 

n.Jf  =  ,.g„. 

III.    Ca^du=~lU- 
J  log  a 


212  INTEGRAL   CALCULUS 

IV.     Ceudu  =  eu. 

V.  I  cos  u  du  =  sin  u, 

VI.  I  sin  u  du  =  —  cos  u. 

VII.  I  sec2  udu  =  tan  w. 

VIII.  I  esc2  «*  rfi*  =  —  cot  u, 

IX.  J  sec  u  tan  ^  d«*  =  sec  u. 

X.  I  esc  u  cot  ?^  f/«j  =  —  esc  u. 

XI.  f     rfM      =sin-iM.  or-cos-t^. 

XII.     f    ^M  ,  =  tan-i  w,  or  -  cot-i  t*. 
J  1  +  w2 

114.  Certain  general  principles.  In  applying  the  above  for- 
mulas of  integration  certain  principles  which  follow  from  the 
rules  of  differentiation  should  be  made  use  of. 

(a)  Tlie  integral  of  the  sum  of  a  finite  number  of  functions  is 
equal  to  the  sum  of  the  integrals  of  the  functions  taken  separately. 

This  follows  from  Art.  10. 
For  example, 


I dx  =  I  x dx  —  I  — ■  =  - —  lo 

J       x  *s  J    x        "J 


sx. 


GENERAL  PRINCIPLES  OF  INTEGRATION  213 

(b)  A  constant  factor  may  be  removed  from  one  side  of  the 
sign  of  integration  to  the  other. 

For,  since  d(cu)  =  cdu, 

it  follows  that  I  cdu—cu-c  I  du. 

To  illustrate,  let  it  be  required  to  integrate 

x2  dx. 


/< 


The  numerical  factor  5  is  first  placed  outside  the  sign  of 
integration,  after  which  formula  I  is  applied.     Accordingly, 

n  ^3 


I  5  x2  dx  =  5  I  x2  dx  =  -—. 

Again,  suppose  the  integral 


/, 


x  dx 


is  to  be  found.  We  notice  that  if  the  numerator  had  an  addi- 
tional factor  2,  it  would  be  the  exact  differential  of  the 
denominator,  and  formula  II  would  be  applicable.  All  that  is 
required,  then,  in  order  to  reduce  the  given  integral  to  a  known 
form,  is  to  multiply  inside  the  sign  of  integration  by  2  and 
outside  by  \.     This  gives 

Cxdx       1  C2xdx     lAI(x2  +  l)      li      /oliN 

In  this  connection  it  must  not  be  forgotten  that: 

An  expression  containing  the  variable  of  integration  cannot  be 

moved  from  one  side  of  the  sign  of  integration  to  the  other. 

(c)  An   arbitrary   constant  may   be   added    to    the  result  of 

integration. 


214  INTEGRAL  CALCULUS 

For,  the  derivative  of  a  constant  is  zero  and  hence 

du  =  d(u  +  c), 
from  which  follows 

d(u  +  c)  =  u  +  c. 


fau=f' 


This  constant  is  called  the  constant  of  integration. 

From  the  preceding  remark  it  follows  that  the  result  of 
integration  is  not  unique,  but  that  any  number  of  functions 
(differing  from  each  other,  however,  only  by  an  additive  con- 
stant) can  be  found,  each  of  which  has  the  same  given  expres- 
sion as  its  derivative.     [Compare  Art.  10,  Cor.] 

Thus,  any  one  of  the  functions  x2  —  l,  x*  +  l,  tf  +  a2, 
(x—a)(x-\-a)  may  serve  as  a  solution  of  the  problem  of  inte- 
grating  f  2  x  dx. 

It  often  happens  that  different  methods  of  integration  lead 
to  different  results.  All  such  differences,  however,  can  occur 
only  in  the  constant  terms. 

For  example, 

f 3  (x  + 1)2  dx  =  3 |f(aj  +  l)2  d{x  + 1)  =  (x  +  l)3 

=  aj8  +  3a2-f3a;+l. 

Integration  of  the  terms  separately  gives 

Csx2dx-\-  C$xdx+  ^3^  =  ^  +  3^4-3^, 

a  result  that  agrees  with  the  preceding  except  in  the  constant 
term. 

Again,  from  formula  XII, 
dx 


s 


tan-1  x,  or  —  cot-1  x. 
ar'  +  l 


GENERAL   PRINCIPLES   OF   INTEGRATION  215 

It  does  not  follow  from  this  that  tan_1&  is  equal  to  —  cot-1#. 
But  they  can  differ  at  most  by  an  additive  constant.  In  fact, 
it  is  known  from  trigonometry  that 

—  cot-1  x  =  tan-1  x  -f  for  +  j , 

_ 

in  which  k  is  any  integer. 

In  a  similar  manner  the  different  results  in  formula  XI  can 
be  explained. 

EXERCISES 

Integrate  the  following : 

1.     (Vxdx.  11.     fCSc2j:^. 

J  J    cot  X 

[Hint.     For  the  purpose  of  in- 


tegration this  may  be  written  12.     f  -sin 

(x%  dx.~] 


J  ~  dx 

fw*.  13.    f    *    L/      * 

J  J  x  log  x\_    t/    log  x 


+  cos  a; 

r 


2.     \  xa  dx. 
3 


log  x  L    «'     log  a: 

f  ,/.*  14  C5x2dx 

'      J      3-'  '  J  X*  +  l    ' 

V  x 

4     Cmt*JJzm  15.  Jtana:^[=-|-^s^]. 

5.    jV -**)•**.  16  ^otxdx. 

6     C5x*-^x  +  ldx  17  Ceaxdx, 

7.  (x(x2+  a2)'2dx.  H8.  (e**xdx. 

8.  J(aar+6)»rfar.  *  19.  f  («-!-&)  »»+n*a*a:. 


20.    I  cos  2  xdx. 


9'    J^i-  20-  J' 

,n      C(n  -x)dx  r  . 

u"     J  Ti 7  *  21.     i  sin  nz  rt.r. 

^  2  ax  -  x1  J 


216  INTEGRAL   CALCULUS 

,  22.    fco**xdx[=P+c™2xdx~]. 

23-    \sln2xdx.       i24.    i  sin(m  -f  n)x  dx.         25.    Kxsmx^dx. 

26.     (  co$?xdx\  =  i  (1  —  sin2a:)cos  xdx   .  27.    (  sin3a;G?a;. 

28.    rtan2xr/xT=  f(sec2a;  -  l)r/af|.  29.    f tan2 x sec2 ztf* 

30.    \  esc2  (ax -\- b)dx.  31.    (  Vcot  a:  •  esc2  x dx. 

32      f         ^r         f  —  rsec2.rr/.r~j 
•/  sin  x  cos  a;  L     J     tan  a:   J 

33.     \  sec3  a:  tan  a;  dx. 

34.  rtnTirr/r.  35.  r 

J     sec  a-  ./ 


r/s 


Va2  -  x2 
[Hint.     Divide  numerator  and  denominator  by  a  and  then  write 

in  the  form 


J  ^w 


36. 


J 


dx 


Vl  -  4  x2 


37. 


J  a2  +  m5 


38.    f-^- 
J  a2a-2  + 


fta 


39.  f ** r=f  ^-2>  i. 

J  x-2  _  4  3  +  5  L     J  (a:  -  2)2  +  1 J 


115.   Integration  by  parts.     If  u  and  v  are  functions  of  x,  the 
rule  for  differentiating  a  product  gives 

d(uv)  =  v  du  -J-  u  dv, 
whence,  by  integrating  and  transposing  terms,  we  have 
I  u  dv  —  uv  —  |  v  du. 


GENERAL  PRINCIPLES   OF  INTEGRATION  217 

This  formula  affords  a  most  valuable  method  of  integration, 
known  as  integration  by  parts.  By  its  use  a  given  integral  is 
made  to  depend  on  another  integral,  which  in  many  cases  is 
of  a  simpler  form  and  more  readily  integrable  than  the 
original  one. 

Ex.  1.  J  \ogxdx. 

Assume  u  =  log  x,   dv  =  dx. 

Then  du  =  — ,    v  =  x. 

x 

By  substituting  in  the  formula  for  integration  by  parts,  we  obtain, 

i  log  xdx  —  x  log  x  —  \  dx 

=  x  log  x  —  x  =  xQog  x  —  1 ) 

=  x(\og  x  —  log  e)  =  x  log  - . 
e 

Ex.  2.  (xexdx. 

Assume  u  =  x,   dv  =  e*  dx. 

Then  du  =  dx,    v  =  ex, 

and  (  xex dx  =  xex  —  (  ex dx  =  e?(x  —  1). 

Suppose  that  a  different  choice  had  been  made  for  u  and  dv  in  the 
present  problem,  say  u  =  ^   rfy  =  ^ 

From  this  would  follow 

r2 

du  —  ex  dx.    v  =  — , 
2 

and  j  xex  dx  =  |  x2ex  —   I  —exdx. 

C x2 
It  will  be  observed  that  the  new  integral  j  —e*dx  is  less  simple  in 

form  than  the  original  one ;    hence  the  present  choice  of  u  and  dv 
is  not  a  fortunate  one. 

No  general  rule  can  be  laid  down  for  the  selection  of  u  and  dv. 
Several  trials  may  be  necessary  before  a  suitable  one  can  be  found. 


218  INTEGRAL   CALCULUS 

It  is  to  be  remarked,  however,  that  dv  should  be  so  chosen  that  its 
integral  may  be  as  simple  as  possible,  while  u  should  be  so  chosen 
that  in  differentiating  it  a  material  simplification  is  brought  about. 
Thus  in  Ex.  1,  by  taking  u  —  log  x,  the  transcendental  function  is 
made  to  disappear  by  differentiation.  In  Ex.  2,  the  presence  of  either 
x  or  ex  prevents  direct  integration.  The  first  factor  x  can  be  removed 
by  differentiation,  and  thus  the  choice  u  =  x  is  naturally  suggested. 


Ex.  3. 


(  x2axdx. 


From  the  preceding  remark  it  is  evident  that  the  only  choice  which 
will  simplify  the  integral  is 

u  =  x2,        dv  =  axdx. 


a 

Hence  du  =  2  xdx,    v  =  - , 

log:  a 


and 


(x2a*dx  =  -^- —  (xaxdx. 

J  lotr  a      log:  a J 


Apply  the  same  method  to  the  new  integral,  assuming 
u  =  x,   di'  =  axdx, 


whence  du  =  dx,   v  — , 

log  a 

and  f  xax  dx  =  ^-  -  — !—  f  ax  dx 

■J  log  a      log  a  J 


x(t  ar 


log  a      (log a)'2 
By  substituting  in  the  preceding  formula,  we  have 

J  log  a  L  log  a      (logo)2  J 


GENERAL  PRINCIPLES  OF   INTEGRATION  219 


EXERCISES 

1. 

i  sin~ixdx. 

7. 

i  x  cot-1  x  dx. 

2. 

f  «*  tan"1  (€*)</*. 

8. 

\  x  sin  3  x  dx. 

3. 

1  a:2  cos  xdx. 

9. 

\  e*  cos  x  dx. 

4. 

\xn\ogxdx. 

10. 

\  ex  sin  xdx. 

5. 

1  a*2  tan-1  xdx. 

11. 

\  cos  a:  cos  2  a:  /7a*. 

6. 

f  sec  a:  tan  x  log  cos  a: 

dx. 

12. 

I  x  sec2  a-  </a\ 

116.  Integration  by  substitution.  It  is  often  necessary  to 
simplify  a  given  differential  f'(x)clx  by  the  introduction  of  a 
new  variable  before  integration  can  be  effected.  Except  for 
certain  special  classes  of  differentials  (see,  for  example,  Arts. 
127-129)  no  general  rule  can  be  laid  down  for  the  guidance  of 
the  student  in  the  use  of  this  method,  but  some  aid  may  be 
derived  from  the  hints  contained  in  the  problems  which  follow. 

Ex.  1.    |     xdx     . 


Va2  -  x2 

Introduce  a  new  variable  z  by  means  of  the  substitution  a-—  x2=  z. 
Differentiate  and  divide  by  -  2,  whence  xdx—  —  — .     Accordingly, 

The  details  required  in  carrying  out  this  substitution  are  so  simple 
that  they  can  be  omitted  and  the  solution  of  the  problem  will  then 
take  the  following  form  : 

f     XdX      =  (V  -  **)"*  xdx  =  -  I  ( («2  -  x*y*(  -  2  xdx) 
J  Vf/2  -  a-2      J  •  -J 


220  INTEGRAL   CALCULUS 

In  this  series  of  steps  the  last  integral  is  obtained  by  multiplying 
inside  the  sign  of  integration  by  —  2  and  outside  by  —  I,  the  object 
being  to  make  the  second  factor  the  differential  of  a2  —  x2.  Think- 
ing of  the  latter  as  a  new  variable,  the  integrand  contains  this 
variable  affected  by  an  exponent  (-  i)  and  multiplied  by  the  differ- 
ential of  the  variable,  in  which  case  formula  I  can  be  applied. 

Ex.  2.    r !°£*  dx. 
J     x 

Assume  log  x  =  z.. 

Then  --  =  dz, 

x 

and  fl^?^=f^  =  l2  =  -Q^. 

J     x  J  2  2 

Here  again  it  is  not  necessary  to  write  out  the  details  of  the  sub- 
stitution, as  it  is  easy  to  think  of  log  a;  as  a  new  independent  variable 
and  to  perform  the  integration  with  respect  to  it.  It  is  then  readily 
seen  that   the   expression  to  be  integrated  consists  of   the  variable 

dx 
log  #  multiplied  by  its  differential   — :,  and  that   the  integration  is 

x 
accordingly  reduced  to  an  immediate  application  of  the  first  formula 

of  integration.     Thus 

k2 


ij\ogx.d(\ogx)  =  Off*)' 


Ex.  3.    j 


Ex.  3.     \  eian    * 


dx 


gives 


1  +  x2 
Think  of  tan-1  x  as  a  new  variable  and  apply  formula  IV.     This 

retan-'*_^    _   fgtan-'x, /(tan-la-)  =  ^tan^x. 

J  1  +  x2      J 


Ex.  4.    C*™-1***. 

dx 


j- 


Ilegard  sin-1  a:  as  a  new  variable  and  -  as  the  differential 

Vl  —  x2 
of  that  variable.     Apply  formula  I. 


GENERAL   PRINCIPLES   OF  INTEGRATION  221 

Ex.  5.     j*02  +  2  x  +  3)0  +  l)dx. 

Multiply  and  divide  by  2.     The  integral  then  takes  the  form 
If  (a?  +  2  x  +  3)  •  (2  x  +  2)rfar. 

Observing  that  (2  x  -\-  2)dx  is  the  differential  of  x2  +  2  x  +  3,  and 
using  the  latter  expression  as  a  new  variable,  we  see  that  formula 
I  is  directly  applicable,  leading  to  the  result 

Ex.  6.      flog  cos  (x2  +  1)  sin  (a;2  +  1)  •  xdx. 

Make  the  substitution         x"2  +  1  =  z. 
The  given  integral  takes  the  form 

-  I  log  cos  z  sin  z  dz. 

Make  a  second  change  of  variable, 

cos  z  —  y. 
Then  sin  zdz  =  —  dy. 

The  transformed  integral  is 

-lijlogydy, 

to  which  the  result  of  Ex.  1,  p.  217,  can  be  at  once  applied. 

It  will  be  observed  that  two  substitutions  which  naturally  suggest 
themselves  from  the  form  of  the  integrand  are  made  in  succession. 
The  two  together  are  obviously  equivalent  to  the  one  transformation, 

cos(x2  +  1)  =  y. 

Ex.7,      f       du       ■  Ex.8,      f     du     . 

J  y/cfi  _  u*  J  u-  +  a2 

-[Hint.     Substitute  u  =  as.] 


222 

INTEGRAL  CALCULUS 

Ex.  9.   f 

dx 

/„.9 

—  •          Ex.  10. 

^ 


dx 


V2  ax  —  x* 
Hint.     Substitute  x  =  -  •  [Hint.     Substitute  x  =  z  +  a.] 

Ex.  11.      |  esc  m  du. 

Multiply  and  divide  the  integrand  by  esc  u  —  cot  u.     It  will  then 

be  seen  that  the  integral  has  the  form   i  — 

J  z 

Another  method  would  be  to  use  the  trigonometric  formula 

n    •     u        u 
sin  u  =  2  sin  -cos  -, 
2        2  \ 

2    V2/       Cdt 
=1—1 

tan^  J  l 

2 


Ex.  12. 


\  sec  u  du. 


Put  u  =  z and  use  Ex.  11. 

2 

Solve  the  problem  also  by  means  of  substitutions  similar  to  those 
used  in  the  preceding  example. 

sin3x 


Ex.13.      (x*y/a*-x*dx.  Ex.15.      f£°l£ 
J  J     sin- 

Ex.14,      f    *a'7ar    •  Ex.16,      f 
J(a;-1)3  J 


dx 


cos2  a;  -f  2  sin2  x 
Put  tan  x  —  z. 

Ex.  17.   Prove  that  I  — — can  be  integrated  by  a  substitution, 

J  (a  +  bx)n 

when  m  is  a  positive  integer. 

117.  Additional  standard  forms.  The  integrals  in  Exs.  7,  8, 
11,  12  of  the  preceding  article,  and  in  Exs.  15,  16  of  Art.  114, 
are  of  such  frequent  occurrence  that  it  is  desirable  to  collect 


GENERAL   PRINCIPLES   OF   INTEGRATION  223 

the  results  of  integration  into  an  additional  list  of  standard 
forms.  Two  other  very  useful  formulas  are  also  included,  the 
derivation  of  which  we  now  give. 

du 


Integration  of  f     du 


Make  the  substitution 


u  +  Vw2  4- «  =  z. 
From  this  equation,  we  obtain,  by  differentiation, 

(l  +       U       }  du  =  dz ; 
\        Vw2  +  a  J 


du 


that  is,  ( VV  +  a  +  u)  — — =  =  dz, 

Vw*  +  a 


whence, 


du  dz  dz 


Vm2  +  a      Vw2  +  a-\-u      z 
This  gives,  on  integrating, 

>_*    ,r*_log, 

J  Vm2+«     ^    z 


=  \og(u  -f  Vw2  +  a). 

Integration  of  f   fu  o  • 

J  w2  —  a2 

The   fraction  — may   be   written  as   the  sum  of   two 

u2  —  a" 


simpler  fractions, 

_l_=j_[^ l_i 

u-  —  a2     2a\_u  —a      u  +  a  J 


224  INTEGRAL   CALCULUS 

whose  denominators  are  the  factors  of  u2  —  a2.     Hence, 

r    du     _  1     rr_du_        du   1 
J  u2  —  a2     2  a  J  [_u  —  a      u  -f  a  J 


li      /  x  .1       1    ,      u—  a 

J  a        it  4-  ct 


=  2^  log  (w  -  a)- log  (w 4- a) 


XIII.  du       =Sin-^ 


Va2 


/; 

XIV.     f     **"      -  log  (u  4  VttHo). 


a 


u  —  a 
u  +  a 


XVII.  I  tan  udu  =  -  log  cos  u  -  log  sec  u. 

XVIII.  J  cot  udu  =  log  sin  u. 

XIX.  J  sec  w  du  =  log  (sec  ™  +  tan  u)  =  log  tan  (tt  +  jY 

XX.  J  esc  m  r  Jw  =  log  (esc  u  —  cot  w)  =  log  tan  „  • 

118.   Integrals  of  the  forms 

/•(Ax  +  B)dx  and    /»  (^g;  +  B)dx 
ax2  +  bx  +  c         J  ■y/ax*+  bx  +  c 

Such  integrals  occur  so  frequently  that  they  deserve  special 
mention.  The  integration  is  facilitated  by  the  substitution 
of  a  new  variable  t  which  reduces  the  affected  quadratic 
ax2  4-  bx  4-  c  to  a  pure  quadratic  of  the  form  mt2  +  n.  The 
mode  of  procedure  will  be  readily  understood  from  the  follow- 
ing illustrative  problems. 


. 


GENERAL   PRINCIPLES   OF   INTEGRATION  225 

x,       ,                                  C          xdx 
I,x.  1.  \ — -. 

J  2  x2  +  2  x  +  3 

The  first  step  is  to  complete  the  square  of  the  x  terms  in  the 
denominator.  After  the  factor  2  has  been  placed  outside  the  integral 
sign,  the  quadratic  expression  may  be  written 

(x*  +  x  + 1)  +  a  -  \)  =  (x  +  \y  + 1 . 

Now  substitute  a  new  variable  t  in  place  of  x+\.     Since  x  =  t  —  \  and 
dx  =  dt,  we  obtain  for  the  new  form  of  the  given  integral 


1  r(t-\)dt  =  \  r'ltdt      If    dt 


**+* 


1L 


=  ilog(*2+-^ —  tan 


2V5  V5 

Ex.2,     f     (2^-1)^ 

J  VI  +2^-3^' 

Divide  out  V3  from  the  denominator;  since  the  coefficient  of  x2  is 
negative,  put  the  x  terms  in  parentheses  preceded  by  the  negative  sign 
and  complete  the  square.     The  integral  then  becomes 

\)dx. 


V'SJ  v'|  -  (x  -  i)* 

Now  make  the  substitution  x  —  \=  t.     Since  dx  =  <//,  the  integral 
reduces  to 

V(-^=-^(M4(--   -^- 


V.jJ    V|  -  **  ViiJ  \9  /     3V3J  V-f  -  t'2 

2 


'3A9        ;        3V3  V  2  / 

2     /r~2  I         1      •      J3x-1\ 

-v/-  -f  -  :r  —  a;2 sin-1    - 

Vo-^      3  3V3  V      2      ; 

-  -a/1  +  2  a:  -  3  x2  -  —  siu-*f  3  *  ~  *  V 
3^  3V3  V      2      J 


EL.    CALC  —  15 


226  INTEGRAL   CALCULUS 

It  is  seen  from  the  two  preceding  examples  that  the  met) 
here  used  contains  two  essential  steps : 

(1)  Completing  the  square  of  the  x  terms  in  ax2  -\-  bx  +  c  ; 

(2)  Substituting  a  new  variable  for  the  part  in  parentheses. 

If  the  numerator  of  the  new  integral  contains  two  terms, 
separate  into  two  integrals  and  integrate  each  one  separately. 

EXERCISES 

1    C         dx  8    C  ('2x~  3)rix 

J3x2-2x  +  5'  9.    )\}~X  dx. 


dx 


J  8  +  4  x  -  4  x2 

4 

r          dx 

^  V30  x  -  9  xi  -  24 

5. 

C          x  dx 

*  Vx*  +  2  x  +  2 

6. 

J  VI  +  2  x  -  x2 

7 

C     (4x  +  5)dx 

*  V8  -  4  x  -  4  x2 

119.    Integrals  of  the  forms 

+ 

[Rationalize  the  numerator.] 
1Q     r(3x  +  2)rfz, 

11. 


]•# 


12  f       (2  x _+_l),/a       . 
J  V-2^-  3z-  1 

13  r       (x-Z)dx 

'  J  V-3s2-2a;  +  l 


f-  **  and    f 


dx 


(Ax  +  ByWax*  +  bx  +  c 


Integrals  of  these  types  can  be  reduced  to  forms  given  in 
the  preceding  article  by  means  of  the  reciprocal  substitution 


GENERAL   PRINCIPLES   OF   INTEGRATION  227 

I  x  EXERCISES 

C       dx  7     f_. dx 

*  xVx*  +  a2  ^  (x  +  2)  V- a:2  -  10x-7 

8.    f         dx         . 
J  ar2Va2  —  x2 


^ 


6 


xVx2  +  a2 

rfx 

xVa2  —  x2 

dx 

x  Vo  x-  —  4  x  + 

1 

rte 

(x  +  l)Vx2  +  2 

X  +  3 

tfs 

(x  +  l)Vx2  +  x 

+  1 

tfx 

^  x2v  x2  -  a2 


10.  f_ 


rfj; 


5.    f-  dx  '  JxWx*  +  a* 

r  ^/x  ii  r        -  fix 

*  (\  -  x^  \f'2  x2  -  4  r.  +  1  ^  f2  x  - 


(1  -  x)  v/2  x1  -  4  x  +  1  J  (2  x-  -  1)  V4  x'2  -  3 


EXERCISES   ON   CHAPTER   I 

1.     (  e**  ex  dx.  7.     \x(a'2  -  x2)*  dx. 

8.  f      *•«      ■ 

9.  f     •" ■ 

J   y/X  +  1   +  Vx  -   1 

10.  i  cos  8  x  dx. 

11.  (  sec  3  x  dx. 

12-     1  fr  sin  e*  rfx. 


2. 

f  5  xzdx 

J  4  +  x8 

3 

f   (2  +  3x2)f/x 

J  6  x3  +  12  x  +  5 

4. 

f1  +  *rf*. 

J    Vx 

5. 

f       *       . 

J  ^8  -  2  « 

6 

r     //x 

228  INTEGRAL   CALCULUS 

dx 


13 


C   sin  x  dx  23      f_ 

J  a  cos  x  +  b  *  6xVl  -  log  a; 

C      dx 

'     1     ,  «/.      C    exdx 

J  Vl  -  e2*  24.     \ -• 


14 

.  [Put  e—  =  «.] 
15 


r  s<fo  .  25.  j* 


ex  +  e- 

cos  0  dO 


y/l  _  ^4  vl  +  cos'2  0  -  sin  ^ 


16. 


S  ,      dX  ■  26.    f- 


dx 


V'4  a:4  +  8  a;2  J  a:  (log  a:)2  +  x 

17     leT^-  27.     f(       seC*      Vd,. 

e       e  J  \a  -  b  tan  x/ 


18.     \%4  tan-1  a;  da;. 


28.  J. 


(a:  —  a)  dx 


Cx~  dx  J  Va*-a2(x  —  a)2—(x-a)4 

J    ax 


2,j_ 


™      f       ^0       r      fl-sin^./n 

20- )m^L=3^s^rfe]-      "«-v8^+2«+i 

._      f    tan  (9^  r 

Ja  +  fttan'0"  31.    J  sin  a:  log  tan  x  dx. 

32  r    ^    r—  f  s*n  x ^x 

J  1  +  cot  a:  L     J  sin  a:  4-  cos  x 

—  1  f  (sm  x  +  cos  x)  —  (cos  x  —  sin  a:)  , 
2  J  sin  a;  +  cos  x 

=  1  rA.cosar-sinarX^l 

2  J  V        sin  a;  -f  cos  a;/      J 

[Another  method  would  be  to  multiply  numerator  and  denominator 
by  sin  a:(cos  x  —  sin  x)  and  express  in  terms  of  the  double  angle.] 


CHAPTER   II 

REDUCTION  FORMULAS 

120.    In  Arts.  118,  119  the  integration  of  certain  simple  ex- 


pressions containing  an  irrationality  of  the  form  ^/atf+bx+c 
was  explained.  As  was  shown  in  Art.  118,  the  radical  can 
be  reduced  to  the  form  V  ±  x2  ±  a?  by  a  change  of  variable. 
It  remains  to  show  how  the  integration  can  be  performed  in 
in  such  cases  as,  for  example, 

xndx 


j  zw  V  ±  x2  ±  a2dx,  J  - 


V  ±  a?  ±  ci- 
ri  being  any  integer. 

For  this  purpose  it  is  convenient  to  consider  a  more  general 
type  of  integral  of  which  the  preceding  are  special  cases,  viz., 

Car  (a  +  bxn)pdx,  (1) 

in  which  m,  n,  p  are  any  numbers  whatever,  integral  or  frac- 
tional, positive  or-negative. 

It  is  t$>  be  remarked  in  the  first  place  that  n  can,  without 
loss  of  generality,  be  regarded  as  positive.  For,  if  n  were 
negative,  say  n  =  —  n',  the  integrand  could  be  written 

xmfa  +  IV  =  xmfaxn'j'  5V  =  xm-*n'(b  +  ax*'y 

This  expression,  which  is  of  the  same  type  as  xm(a  +  bxn)p,  is 
such  that  the  exponent  of  x  inside  the  parentheses  is  positive. 

229 


230  INTEGRAL   CALCULUS 

It  will  now  be  proved  that  an  integral  of  the  type  (1)  can  in 
general  be  reduced  to  one  of  the  four  integrals 

(a)  A  j  xm~n(a  +  bxn)pdx,  (b    Ai  xm+n(a  +  bxn)pdx, 

(c)  A  I  xm(a  +  bxn)p-*dx,  (d)  A  (  xm(a  +  bxn]*+ldxf 

plus  an  algebraic  term  of  the  form 

BxK(a  +  bxny. 

Here  A,  B,  \,  fx  are  certain  constants  which  will  be  deter- 
mined presently. 

Observe  that  in  each  of  the  four  cases  the  integral  to  which 
(1)  is  reduced  is  of  the  same  type  as  (1),  but  that  certain 
changes  have  taken  place  in  the  exponents,  viz., 

the  exponent  m  of  the  monomial  factor  is  increased  or  dimin- 
ished by  n, 

or,  the  exponent  p  of  the  binomial  is  increased  or  dimin- 
ished by  unity. 

The  values  of  A.  and  fi  are  determined  by  the  following  rule  : 

Compare  the  exponents  of  the  monomial  factors  in  the  given 

integral  and  in  the  integral  to  which  it  is  to  be  reduced.     Select 

the  less  of  the  two  members  and  increase  it  by  unity.      The  result 

is  the  value  of\.     In  like  manner,  compare  the  exponents  of  the 

* 
binomial  factors  in  the  two  integrals,  select  the  less,  and  increase 

it  by  unity.     This  gives  /x. 

Thus,  if  it  is  desired  to  reduce  the  given  integral  to 

A  j  xm~n(a  4-  bxn)pdx, 

first  write  down  the  formula 

j  xm(a  +  bxn)pdx  —  A  \  xm~n(a  4-  bxn)pdx  +  Bxx(a  4-  &»■)*. 


REDUCTION   FORMULAS  231 

The  exponents  of  the  monomial  factors  in  the  two  integrals 
are  m  and  m  —  n  respectively,  of  which  m  —  n  is  the  less. 
This,  increased  by  unity,  gives  the  value  of  X;  that  is, 
X  =  m  —  n  +  1. 

Again,  the  exponent  of  the  binomial  factor  in  each  integral 
is  the  same,  namely  p,  so  that  there  is  no  choice  as  to  which  of 
the  two  is  the  less.  Increase  this  number  p  by  unity  to  obtain 
the  value  of  /x.     Hence  /x  =  p  -f  1. 

The  above  formula  may  now  be  written 

J  xm(a  -(-  bxn)pdx 

=  A  Cxm-'l(a  +  bxn)pdx  +  Bxm-n+1(a  +  bxn)p+\       (2) 

In  order  to  determine  the  values  of  the  unknown  constants 
A  and  B,  simplify  the  equation  by  differentiating  both  mem- 
bers. After  being  divided  by  xm~n(a  +  bxn)p  the  resulting 
equation  is  reduced  to 

xn  =  A  +  Ba(m  —  n  + 1)  +  Bb(m  +  np  +  l)a,-n. 

By  equating  coefficients  of  like  powers  of  x  in  both  members, 
we  find  the  values  of  A  and  B  to  be 

A  _       a(m  -  m  + 1)       B=  1 

b(m  +  np  -h  1) '  b(m  +  np  +  l)' 

When  these  values  are  substituted  in  formula  (2),  it  becomes 

J  xm(a  +bxn)pdx 


ci{m  —  n  + 1) 


+  np+l)J  v  7  6(m+np  +  l)       LJ 

Notice  that  the  existence  of  formula  (2)  has  been  proved 
by  showing  that  values  can  be  found  for  A  and  B  which  make 
the  two  members  of  this  equation  identical. 


232  INTEGRAL   CALCULUS 

There   is   one   case,   however,  in  which   this   reduction   is 

impossible,  viz.,  when 

m  +  np  +  1  =  0, 

for  in  that  case  A  and  B  become  infinite.     [See  Ex.  4,  p.  235.] 
In  a  similar  manner  the  three  following  formulas  may  be 
derived : 


J  xm(a  +  bxn)p  dx 


a(m  +  l)       J         k  -r      j  a(m  +  l) 

I  £m(a  +  bxn)pdx 

: l I  xm(a  +  bxn)p  xdx  H *■—-*- f-  •  TC] 

m  +  np  + 1  •/  m  +  ?ip  +  1 

I  xm(a-\-bxn)pdx 

:  — ! — i — —  I  xm(a  +  bxn)p+hlx i — ■ - TDl 

an(j>  +  l)      J  awQj  +  1) 

The  cases  in  which  the  above  reductions  are  impossible  are, 
For  formulas  [A]  and  [C],  when  m  +  np  +  1  =  0; 
for  formula  [B]  ,  when  m  +  1  =  0 ; 

for  formula  [D]  ,  when  _p  +  1  =  0. 


..     (xsVa 


Ex.1.     \  x8Va*  -  x*  <lx. 


If  the  monomial  factor  were  x  instead  of  x3,  the  integration  could 
easily  be  effected  by  using  formula  I.  Since  in  the  present  case 
m  —  3,  n  =  2,  formula  [A],  which  diminishes  in  by  n,  will  reduce  the 
above  integral  to  one  that  can  be  directly  integrated. 


REDUCTION   FORMULAS  233 

Instead  of  substituting  in  [A],  as  might  readily  be  done,  it  is  best 
to  apply  to  particular  problems  the  same  mode  of  procedure  that  was 
used  in  deriving  the  general  formula.  There  are  two  advantages  in 
this.  First,  it  makes  the  student  independent  of  the  formulas,  and 
second,  when  several  reductions  have  to  be  made  in  the  same  problem, 
the  work  is  generally  shorter.      [See  Ex.  4.] 

Accordingly  assume 

f  x\d2  -  xrf  dx  =  A  ( x(a2  -  x2y  dx  +  Bx2(a2  -  x2)\ 

the  values  of  X  and  ft  having  been  determined  by  the  previously  given 

rule. 

Differentiate,  and   divide   the  resulting  equation   by  x(a2  —  x2y. 

This  gives  . 

6  x2  =  A  +  B(2a2-5x2), 

from  which,  on  equating  coefficients  of  like  powers  of  x, 

o  5 

hence, 

(x3^d2-x2dx  =  —( (a2  ~  x2) * xdx-\  x\a2  -  z2)* 
=  -  ^(2  a2  +  3x2)  (a*-xrf. 


:.  2.     (Vx2 


Ex.  2.     \  Vx2  -2x  -6  dx. 


By  following  the  suggestions  of  Art.  118,  this  integral  can  be  re- 
duced to  the  form 

C  Vz2  -  4  dz, 


in  which  z  =  x  —  1. 
Assume 


jV  -  4)Klz  =  A§(z2  -  4)~^  dz  +  Bz(z2  -  4)i 

In   determining   A   notice   that  m  =  0  in  both  integrals,  so  that 
X  =  0  +  1  =  1.     Also,  fji  =  -  £  +  !  =  £. 


234  INTEGRAL   CALCULUS 

Ex.3.    (  V2  ax  -  xl  dx. 

The  mode  of  procedure  of  Ex.  2  may  be  followed.     Another  method 
can  also  be  used,  as  follows. 
On  writing  in  the  form 

fa* (2  a  -xy  dx, 

and  observing  that  the  integration  of 

jV*  (2  <z -aO"~*tf*=  J 


V2  ax  -  x2 

can  be  performed  (see  Ex.  10,  p.  222),  it  will  be  seen  that  integration 
may  be  effected  in  the  present  case  by  reducing  each  of  the  exponents 
m  and  p  by  unity.  This  is  possible  since  n  =  1  and  m  can  accordingly 
be  diminished  by  1.     Hence  assume 

(V  (2  a-  xy  dx  =  A '  jV  *  (2  a-  x)*  dx  +  B'x?  (2  a-  x)% . 

The  exponent  of  the  binomial  in  the  new  integral  may  be  reduced 
in  turn  by  assuming 

f  aT*  (2  a-  a?)*  dx  =  A"  ( x~?  (2  a-  x)~^  dx  +  B"  x?(2a-x)K 

When  this  expression  is  substituted  for  the  integral  in  the  second 
member  of  the  preceding  equation,  the  result  takes  the  form 

(  V2  ax  -  x2  dx  =  A  (         dx         +  Bx*(2  a  -  a:)*  +  Cx?(2  a  -  xf, 

J  .  a/2  ax  -  x1 

in  which  A,  B,  C  are  written  for  brevity  in  the  place  of  A' A",  A'B", 
B'  respectively.  The  values  of  A,  B,  C  are  calculated  in  the  usual 
manner  by  differentiating,  simplifying,  and  equating  coefficients  of 
like  powers  of  x. 

The  method  just  given  requires  two  reductions,  and  hence  is  less 
suitable  than  that  employed  in  Ex.  2,  which  requires  but  one  reduction. 

The  rule  for  determining  the  values  of  A.  and  /x  may  now  be 
advantageously  abbreviated.     Let  m,p  be  the  exponents  of  the 


REDUCTION   FORMULAS  235 

two  factors  in  the  given  integral,  and  m',  p'  the  corresponding 
exponents  in  the  new  integral.  Of  these  two  pairs,  m,  p  and 
m',  p',  one  of  the  numbers  in  the  one  pair  is  less  than  the  cor- 
responding number  in  the  other  pair.  This  fact  will  be  ex- 
pressed briefly  by  saying  that  the  one  pair  is  less  than  the 
other  pair.  With  this  understanding  the  preceding  rule  may 
be  expressed  as  follows ; 

Select  the  less  of  the  two  pairs  of  exponents  m,  p  and  m',  p'. 
Increase  each  number  in  the  pair  selected  by  unity.  Tliis  gives 
the  pair  of  exponents  A.,  /jl. 

Ex.4,     f     *dx     . 

(x2  +  a2)* 

Assume  successively 

( x\x2  +  a2)~~2  dx  =  A'    (x\x2  +  a2)~2  dx  +  B'x\x2  +  a2)"^, 

(x4(x2  +  a2)~*  dx  =  A"  (x2(x2  +  a2)~*  dx  +  B"x*(x*  +  a2)*, 

(x2(x2  +  a2)"*  dx  =  A'"1j(x2  +  a2)-^  tte  +  i?'".r(.r2  +  a2)i 

These  equations  may  be  combined  into  the  single  formula 

(  x\x2  +  a*)~*dx  =  A  fj  (x2  +  a2)~^dx  +  Bx(x2  +  a2y 
+  Cx*(x2  +  a2)*  +  Dx*(x2  +  a2)""i 

The  values  of  the  coefficients  are  found  to  be 

A=-$a*,     B=-,     C  =  -—.     D  =  —- 
2  a2  a2 

Hence 

( x\x2  +  a*)~*  dx  =  ** +%<***  _  3  a-2  log  (X  +  VX2  +  a2). 
J  2Vx2  •+-  a'2 


236  INTEGRAL   CALCULUS 

In  this  example  three  reductions  were  necessary;  first,  a  reduction 
of  type  [X)],  second,  and  third,  a  reduction  of  type  [-4].  Can  these 
reductions  be  taken  in  any  order? 

The  different  possible  arrangements  of  the  order  in  which  these 
three  reductions  might  succeed  each  other  are 

(1)    UL  Ul  [2>]  ;       (2)    01],  [Z>],  [A]  ;      (3)   [Z>],  Ul  [A], 

of  which  number  (3)  was  chosen  in  the  solution  of  the  problem.     Of 

the  other  two  arrangements,  (2)  can  be  used,  but  (1)  cannot.     For, 

after  first  applying  \_A.~]  (which  would  be  done  in  either  case),  the  new 

integral  is  „  _3 

\  x2(a2  +  x2)    J  dx. 

If  \_A~\  were  now  applied  it  would  be  necessary  to  assume 
f  x\a2  +  x2)~i  dx  =  A  (  (a2  +  x2)~i  +  Bx(a2  +  x2)~\. 

This  equation,  when  differentiated  and  simplified,  becomes 

x2  =  A  +  Ba2, 

a  relation  which  it  is  clearly  impossible  to  reduce  to  an  identity  by 
equating  coefficients  of  like  powers  of  x,  since  there  is  no  x2  term  in 
the  right  member  to  correspond  with  the  one  in  the  left  member.  It 
will  be  observed  that  this  is  the  exceptional  case  mentioned  on  page 
232,  in  which  m  +  np  +  1  =  0. 


EXERCISES 


1.  f  (a2  -  x2y2  dx.  5.     (  Va2  - 

2.  f— ^ 6.    f_ 

J(>2 +  4)2  J  x* 


x2  dx. 
dx 


3.     f ** •  7      C—** 


(x2+a) 


.      C     x2  dx  /»  3 


REDUCTION  FORMULAS                                237 

9.    (Vx^T^dx.  12.    (       dx       • 

10.  f  a;v2ox  -  x*dx.  13     f ^f 

J  '   J  (y+  +  i  x  +  .3)3 

r  y/'^nx    #2   ,  /•     

11.  J  -^— dx.  i4.   j  VI  -  2  x  -  x2dx. 

15.    Show  that 

f       dx       _          1          f         a?  (9     _o\    C         dx       _~| 

J  02  +  c)»  ~  2  c(>  -  1)  L  (a;2  +  c)»-i  x  K"  U       '  }  J  (x2  +  c)«-U ' 


16.     f_^ 19.    f 


17  "^ 


sin0<70 


(1  +  e  sin20)t 

[Substitute  cos  0  =  z.] 
C    xdx 

J  (*2+7)2'  20.   jWa 


ia  J : — Hh^"  21.  f  («2  -  x2)'2 


2-  s2rfa:. 
dx. 


CHAPTER   III 

INTEGRATION  OF  RATIONAL  FRACTIONS 

121.  Decomposition  of  rational  fractions.  The  object  of  the 
present  chapter  is  to  show  how  to  integrate  fractions  of  the 
form  <f>(x) 

wherein  <f>(x)  and  if/(x)  are  polynomials  in  x. 

The  desired  result  is  accomplished  by  the  method  of  sepa- 
rating the  given  fraction  into  a  sum  of  terms  of  a  simpler 
kind,  and  integrating  term  by  term. 

If  the  degree  of  the  numerator  is  equal  to  or  greater  than 
the  degree  of  the  denominator,  the  indicated  division  can  be 
carried  out  until  a  remainder  is  obtained  which  is  of  lower 
degree  than  the  denominator.  Hence  the  fraction  can  be  re- 
duced to  the  form 

iM=aXn  +  bxn-l+    ...    +./M, 
xP{x)  x];(x) 

in  which  the  degree  of  f(x)  is  less  than  that  of  if/(x). 

As  to  the  remainder  fraction  ^  >  '  .  it  is  to  be  remarked  in 

the  first  place  that  the  methods  of  the  preceding  articles  are 
sufficient  to  effect  the  integration  of  such  simple  fractions  as 

A  A'        _.  Mx  +  n    M'x+N'        .      P^±_Q      _    (1) 

x—a(x  —  a)'2'       '    x,2±a2'    (#2±a2)2'       '  x2+mx+ri 

Now  the  sum  of  several  such  fractions  is  a  fraction  of  the 
kind   under   consideration,   viz.,  one   whose   numerator  is   of 

238 


INTEGRATION   OF   RATIONAL   FRACTIONS  239 

lower   degree  than  its  denominator.     The  question  naturally 
arises  as  to  whether   the  converse  is  possible,  that  is :    Can 


f(x) 
every  fraction  ^-^  be  separated  into  a  sum  of  fractions  of  as 

simple  types  as  those  given  in  (1)? 

The  answer  is,  yes. 

Since  the  sum  of  several  fractions  has  for  its  denominator 
the  least  common  multiple  of  the  several  denominators,  it  fol- 
lows that  if  -7^-  can  be  separated  into  a  sum  of  simpler  frac- 

tions,  the  denominators  of  these  fractions  must  be  divisors  of 
$(x).  Now  it  is  known  from  Algebra  that  every  polynomial 
xf/(x)  having  real  coefficients  (and  only  those  having  real  coeffi- 
cients are  to  be  considered  in  what  follows)  is  the  product 
of  factors  of  either  the  first  or  the  second  degree,  the  coefficients 
of  each  factor  being  real. 

This  fact  naturally  leads  to  the  discussion  of  four  different 
cases. 

I.  When  if/(x)  can  be  separated  into  real  factors  of  the 
first  degree,  no  two  alike. 

E.g.,  xf,(x)  =  (x-a)  (x  -  b)  (x  -  c). 

II.  When  the  real  factors  are  all  of  the  first  degree,  some 
of  which  are  repeated. 

E.g.,  if;  (x)  =  (x-a)(x-  b)2  (x  -  cf. 

III.  When  some  of  the  factors  are  necessarily  of  the  sec- 
ond degree,  but  no  two  such  are  alike. 

E.g.,  +  (x)  =  (x2  +  a2)  (x2  +  x  +  1)  (x  -  6)  (x  -  c)2. 

IV.  When  second  degree  factors  occur,  some  of  which  are 
repeated. 

E.g.,  t(x)  =  (x2  +  a2)2(x2-x  +  l)(x-b).    ■ 


240  INTEGRAL   CALCULUS 

122.    Case   I.      Factors   of    the   first   degree,   none  repeated. 

When  ij/(x)  is  of  the  form 

ijz  (xj  =  {x  —  a)  (x  —  b)(x  —  c)  •  •  •  (x  —  n), 

f(x)         A     .      B  G     .  N 

assume  J-\J-  = f- H h  •  •  •  H > 

j/a(.t)      a;  —  a      x  —  b      x  —  c  £  —  ft 

in  which  .4,  5,  C,  •••,  N  are  constants  whose  values  are  to  be 
determined  by  the  condition  that  the  sum  of  the  terms  in  the 
right-hand  member  shall  be  identical  with  the  left-hand 
member. 

Ex.     C*-**  +  *dz. 
J  x\  -  3  x  +  2 

Dividing  numerator  by  denominator,  we  obtain 
x3  —  3  x"2  -f  a:  a: 


.r- 


a;  +  2  x2  -  3  ar  +  2 


Assume         — — ■  =  — - — -  -f 


(x-l)(a:-2)      x-1      x  -  2 

By  clearing  of  fractions,  we  have 
(1)  x  =  A(x  -2)  +B(x-  1). 

In  order  that  the  two  members  of  this  equation  may  be  identical 
it  is  necessary  that  the  coefficients  of  like  powers  of  x  be  the  same  in 
each. 

Hence  1  =  ^+5,     0  =  -  2  A  -  B, 

from  which  A  =  -  1,  B  =  2. 

Accordingly  the  given  integral  becomes 

$(x+^--^dz  =  ^+]og(x-l)-21og(x-2) 

Alog     *-*    . 
2  &  (a;  -  2)2 


INTEGRATION   OF   RATIONAL   FRACTIONS  241 

A  shorter  method  of  calculating  the  coefficients  can  be  used. 
Since  equation  (1)  is  an  identity,  it  is  true  for  all  values  of  x.  By 
giving  x  the  value  1  the  equation  reduces  to  1  =  A(—  1),  or  A  =  —  1. 
Again,  assume  x  =  2.     Whence  2  =  B. 

EXERCISES 
1      r     dx  .     r    (x2  -  ab)  dx 

J  x2  -  a2 '  "   J  (x  -  a)  (x  -  b)  ' 

2.  f1-3*^.  5.  r__^f 

J  xs  -  x  J  x2  -  4  x  +  1 

3      r(x*-V2)dx  6     f_(x2  -  l)dx 

J  x2  +  4  x  +  3 "  '   J  (x* 

?      f  x2  -2cx+flc-  q&  +  & 
J    (x  —  a)(x  —  &)(x  —  c) 

8.     fx2(x  + a)-1^*^)-1^- 

9      r   (3s  +  l)tfar  12      f__^£_ 

J2x2  +  3x-2'  "    J  x2 +  7  a; +  12 

1Q     f     (x2  +  qft)<7x  13     f      tf* 

Jx(x -a)(x+ &)'  '   J  a2x2  -  i2' 


4)(4x2  -  1) 
Jx. 


11      f  ('  +  *)*«  .  14.    f_! 

J  2  x  -  x2  -  x3  J  1 


sec2  x  e/x 


tan*2x 
[Put  tan  x  =  t.~\ 

123.    Case  II.     Factors  of  the  first  degree,  some  repeated. 

Fx      r(ox2-3x+l)^x 
J  x(x  -  l)3 

Assume 
(1)  5  x2  -  3  x  +  1  ^  ,1   [      B      ,         C        ,        D 


x(x-l)3  x      x-1      (x-1)2      (x-1)3 

To  justify  this  assumption,  observe  that: 

(a)  In  adding  the  fractions  in  the  right-hand  member,  the  least 
common  multiple  of   the  denominators  will    be  x(x  —  l)3,  which  is 
identical  with  the  denominator  in  the  left-hand  member. 
el.  calc  — 16 


242  INTEGRAL   CALCULUS 

(6)  Further,  the  expressions  x,  x  —  1,  (x  —  l)2,  (a;  —  l)3  are  the 
only  ones  which  can  be  assumed  as  denominators  of  the  partial 
fractions,  since  these  are  the  only  divisors  of  x  (x  —  l)3  consisting  of 
powers  of  a  prime  factor. 

(c)  When  equation  (1)  is  cleared  of  fractions,  and  the  coefficients 
of  like  powers  of  x  in  both  members  are  equated,  four  equations  are 
obtained,  exactly  the  right  number  from  which  to  determine  the  four 
unknown  constants  A,  B,  C,  D. 

Instead  of  the  method  just  indicated  in  (e)  for  calculating  the 
coefficients,  a  more  rapid  process  would  be  as  follows. 

By  clearing  of  fractions,  the  identity  (1)  may  be  written 

5  x2  -  3  x  +  1=  A(x  -  l)3  +  Bx(x  -  l)2  +  Cx(x  -  1)  +  Dx. 
Putting  x  =  1  gives  at  once     3  =  D. 

Substitute  for  D  the  value  just  found,  and  transpose  the  corre- 
sponding term.     This  gives 

5a;2_  6z  +  1  =  A(x-1)*  +  Bx(x-  1)-+  Cx  (x  -  1). 

It  can  be  seen  by  inspection  that  the  right-hand  member  of  the 
result  is  divisible  by  x  —  1.     As  this  relation  is  an  identity,  it  follows 
that  the  left-hand  member  is  also  divisible  by  x  —  1.      When  this 
factor  is  removed  from  both  members,  the  equation  reduces  to 
5  x  -  1  =  A  (x  -  l)2  +  Bx  (x  -  1)  f  Cx. 

Now  put  x  =  1.     Then  C  =  4. 

Substitute  the  value  found  for  C,  transpose,  and  divide  by  x  —  1. 
The  result  is  1  =  A  (x  -  1)  +  Bx. 

By  giving  x  the  values  0  and  1  in  succession,  we  find  that 

A  =-1,   5=1. 

Accordingly,  we  have 


r(5s»-3ar+l)tf*=  (7      1  +  _L_+        4  »        )dx 

J  X(X-1Y  J\      XX-  (x-1)*       (X- 

=  log; 


x(x-iy        J\   x    x-\    o-i)2    (x-iy 

x-  l        Sx -  5 
2(x  -  l)2' 


INTEGRATION   OF   RATIONAL   FRACTIONS  243 

EXERCISES 

C  dx 4      f(V2r+l)</g 

"  y{x-m*  +  v  '  J^(x+V2)2 

Jx3(x-1)  J     38(3  +   1)3 

T       xr/x  6      C  2(.r3  +  ft2*)*/* 

J  (x2-  a2)2'  '   J  a:4  -  2  a2x2  f  a4' 
7     f             V2^ 


f  — 

J(2  + 


(2  +  V2  -  V2  x)8 

8     rax*  +  n'2x2  +  (a  +  \)x+adx 
J  x2(a  +  x) 

9      r(.r3-  \)dx  ±1      C(x2  -Ux+2G)dx 

J    x3  +3x2'*  'J 


10.     f  (ax2  +  fa8)-1  rfx. 


[Substitute  .r  —  3  =  2.] 
12.     r    x'2<lx 


,       T     .r2  r/x 
J  (x-aY 


(- 

[Substitute  a;  —  a  =  2.] 


124.   Case  III.     Occurrence  of  quadratic  factors,  none  repeated. 

Fx  ±     C     (4x2  +  ox  +  ±)dx 

'  J  (x'2+  l)(x2  +  2x  +  2)' 

Assume 

.jn  4 x2  +  5 a:  +  4  _  J  x  +  i?  Cx  +  /> 


O2  +  l)(x2  +  2  x  +  2)        x2  +  1        x2  +  2  3  +  2 
Then 
(2)  4x2  +  5x+4=(^x  +  £)(z2  +  2a;  +  2)  +  (Ca;  +  Z))(x2  +  1). 

By  equating  coefficients  of  like  powers  of  x 

0  =  A  +  C,  5  =  2 ,4  +  2  5  +  C, 

4  =  2/1+5  +  A  4  =  2/3  + Z), 

from  which  4  =  1,  5  =  2,  C  -  -  1,  Z>  =  0. 

Hence  the  given  integral  becomes 

C(x  +  '2)r/x        C         X'lx  „.  ,  1/       ,    ix,    11  *2+l 

1   —5—1 J  ~ — ; ;  =  2  tan-1^  +  tan"1^  +  l)+£log- — -^ — -. 

J     x2+\         Jx2+2x  +  2  x2  +  2.r  +  2 


244  INTEGRAL   CALCULUS 

To  make  clear  the  reasons  for  the  assumption  which  was  made  con- 
cerning the  form  of  equation  (1),  observe  that  since  the  factors  of  the 
denominator  in  the  left  member  are  x2  +  1  and  x2  +  2  x  -f  2,  these 
must  necessarily  be  the  denominators  in  the  right  member.  Also, 
since  the  numerator  of  the  given  fraction  is  of  lower  degree  than  its 
denominator,  the  numerator  of  each  partial  fraction  must  be  of  lower 
degree  than  its  denominator.  As  the  latter  is  of  the  second  degree  in 
each  case,  the  most  general  form  for  a  numerator  fulfilling  this  re- 
quirement (i.e.,  to  be  of  lower  degree  than  its  denominator)  is  an  ex- 
pression of  the  first  degree  such  as  Ax  +  B,  or  Cx  +  D. 

Notice,  besides,  that  in  equating  the  coefficients  of  like  powers  of  x 
in  opposite  members  of  equation  (2),  four  equations  are.  obtained 
which  exactly  suffice  to  determine  the  four  unknown  coefficients 
A,B,  C,D. 

dx 


Ex.2.    ( 

J  (x2  + 


02+l)02+2) 
We  can  assume  in  this  case - — 


(«+!)(«  +  2) 


(.r2+ l)(a;2  +  2)       x2  +  1      x2  + 
;ion  x2  =  t,  the 
,  to  which  Case  I  is  applicable. 


For  if  we  make  the  substitution  x2  =  t,  the  given  fraction  becomes 
1 


EXERCISES 


±     C    4dx  5     C(ix-6)dx 

J  x8  +  4  x  J    x4  +  2  x'2 

„      C            xdx 6     C       x  dx 

J  (x  +  \){x2  +1)'  J  x*  +  x2  +  1* 

3  f     fix  7      C xdx 

Jx3  +  a8'  '    J  (x  -  a)\x2  +  a2)' 

4  C       (a2-b2)dx  8     C      (x*  +  2x  +  2)dx 

J  (x2  +  a2)(x2  +  b2)'  '  J  (x  -  l)(x2  +  2x  +  2) 

9.    f 2d* 

J  (x-lX^  +  1) 


INTEGRATION   OF   RATIONAL   FRACTIONS  245 

125.  Case  IV.  Occurrence  of  quadratic  factors,  some  repeated. 
This  case  bears  the  same  relation  to  Case  III  that  Case  II 
bears  to  Case  I,  and  an  exactly  analogous  mode  of  procedure  is 
to  be  followed. 


Ex.    f2*5-*4*8*8 +  *,/*. 
J  O2  +  2)3 


Assume 

2x5  -x4  +  8x*  +  i  =  Ax  +  B        Cx  +  D        Ex  +  F 
(x'2  +  2)*  "    x2  +  2        (xa  +  2)a      (^  +  2)3' 

Whence,  by  clearing  of  fractions, 
2x*-x*  +  8xS+4=(Ax+B)(x2  +  2)2+(Cx  +  D)(x2  +  2)+Ex  +  F.     (1) 

Instead  of  equating  coefficients  of  like  powers  of  x,  as  might  be 
done,  we  may  calculate  the  values  of  A,  By  C,  •••  by  the  following 
briefer  method. 

Substitute  for  x2  the  value  —  2,  or,  what  is  the  same  thing,  let 
x  =  V—  2.  This  causes  all  the  terms  of  the  right  member  to  drop 
out  except  the  last  two,  and  equation  (1)  reduces  to 

•  -  8V^2  =  EV^2  +  F. 

By  equating  real  and  imaginary  terms  in  both  members,  we  obtain 
-  8  =  E,  0  =  F. 

Substitute  the  values  found  for  E  and  Fin  (1),  and  transpose  the 
corresponding  terms.  Both  members  will  then  contain  the  factor 
x2  -f  2.     On  striking  this  out  the  equation  reduces  to 

2x*  -x2  +  ±x  +  2  =  (Ax  +  B)(x2  +  2)  +  Cx  +  D. 
Proceed  as  before  by  putting  x2  =  —  2.     Whence 
4  =  CV^2  +  D, 
and  therefore  0  =  C,  4  =  D. 


246  INTEGRAL   CALCULUS 

Substitute  these  values,  transpose,  and  divide  by  x2  -f  2.     This  gives 
2x  -1  -Ax  +  B, 
whence  A  =  2,  i?  =  -  1. 

The  given  integral  accordingly  reduces  to 

J  x2  +  2  J  (x-2  +  2)2       ^  (**  +  2)s 

The  first  term  becomes 

J  x2  +  2      ^  z2  +  2  yo  v  2 

The  second,  integrated  by  the  method  of  reduction  (Chap.  II), 

8"ives                                    x           1    .       .   x 
tan-1 

&  +  2      V2  V2 

Finally,  by  using  formula  I  the  last  term  is  integrated  immediately. 
Hence 

f  2  xb  -  xA  +  8  xs  +  4    ,         ,       /  o   ,   0\    ,        a: 
J  (xl  -4-  2  V  ir-2  -4-  2 


(z2  +  2)3  °  v  x2  +  2      (x-2  +  2)2 

EXERCISES 
+  1/  •>      a:2(ar2+l)s 


J  V^2+  1/  J    x2(x2+  l)2 

2.       f^  +  ^'  +  ^fe.  5.      f^  +  ^'^rft 

J      (x2  -ha2)2  J         (x2  +  a2)2 

3       C  2xdx  6       C     x6  dx 

J  (1+  x)(l+x2)2'  *     J(l  +  x2)8" 

[Ex.  6  can  also  be  integrated,  and  more  easily,  by  means  of  the 
substitution  1  +  x2  =  £.] 

The  principles  used  in  the  preceding  cases  in  the  assump- 
tion of  the  partial  fractions  may  be  summed  up  as  follows : 

Each  of  the  denominators  of  the  partial  fractions  contains  one 
and   only  one  of  the  prime  fac'ors  of  the   given    denominator. 


INTEGRATION   OF   RATIONAL   FRACTIONS  247 

When  a  prime  factor  occurs  to  the  nth  power  in  the  denominator 
of  the  given  fraction,  all  of  its  different  powers  from  the  Jirst  to 
the  nth  must  be  used  as  denominators  of  the  partial  fractions. 

TJie  numerator  of  each  of  the  assumed  fractions  is  of  degree 
one  lower  than  the  degree  of  the  prime  factor  ichose  power  occurs 
in  the  corresponding  denominator. 

126.  General  theorem.  Since  every  rational  fraction  can  be 
integrated  by  first  separating  it,  if  necessary,  into  simpler  frac- 
tions in  accordance  with  some  one  of  the  cases  considered 
above,  the  important  conclusion  is  at  once  deducible : 

Hie  integral  of  every  rational  algebraic  fraction  is  expressible 
in  terms  of  algebraic,  logarithmic,  and  inverse-trigonometric 
functions. 


CHAPTER   IV 

INTEGRATION  BY  RATIONALIZATION 

At  the  end  of  the  preceding  chapter  it  was  remarked  that 
every  rational  algebraic  function  can  be  integrated.  The 
question  as  to  the  possibility  of  integrating  irrational  func- 
tions has  next  to  be  considered.  This  has  already  been 
touched  upon  in  Chapter  II,  where  a  certain  type  of  irrational 
functions  was  treated  by  the  method  of  reduction. 

In  the  present  chapter  it  is  proposed  to  consider  the  sim- 
plest cases  of  irrational  functions,  viz.,  those  containing 
■y/ax  -f-  b  and  yW2  -f  bx  -4-  c,  and  to  show  how,  by  a  process 
of  rationalization,  every  such  function  can  be  integrated. 

127.    Integration    of    functions    containing    the    irrationality 


•y/anc  +  b.  When  the  integrand  contains  -y/ax  +  b,  that  is, 
the  nth  root  of  an  expression  of  the  first  degree  in  x,  but  no 
other  irrationality,  it  can  be  reduced  to  a  rational  form  by 
means  of  the  substitution 


■y/aoc  -\-b  =  z. 

Ex.  1. 

Ki 

dx 

x  +  3  - 

-  1 

Assume 

V2  x  +  13  =  z, 

that  is, 

2  x  +  3  =  z2. 

Then 

dx  =  z  <h, 

and 

f 

dx            _  C  zdz 

z+  log(z-l) 
V2  x  +  3  -  1      sz-i 

=  V2  .rT~3  +  log  (V2T+3  -  1). 

248 


INTEGRATION   BY   RATIONALIZATION  249 


j-l  +  «*r**- 


1  +  X*  —  X1  —  VI 


xz  +  a: 


It  would  appear  at  first  sight  that  this  integrand  contains  several 
irrationalities,  viz.,  Vx,  Vx,  Vx.  It  is  readily  seen,  however,  that 
they  are  all  powers  of  Vx,  and  hence  the  substitution  Vx  =  z  will 
rationalize  the  expression  to  be  integrated. 


EXERCISES 

4. 


5. 


f 

dx 

K 

Vx  +  1 

S; 

fix 

/x+  Vx 

f 

dx 

dx 


(x  -  l)Vx 
dx 


(x  —  a  —  b2)  Vx  —  a 


5- 

2  Vx  -  1  -f  x  x7  +  x 


i  l+^ldx. 


When  two  irrationalities  of  the  form  Vaa  4-  b,  -Vex  -f  d 
occur  in  the  integrand,  the  first  radical  can  be  made  to  dis- 
appear by  the  substitution 


Vax  -\-b  =  z. 
The  second  radical  then  reduces  to 


V 


Ca(z*-b)  +  d, 


and  the  method  of  the  next  article  can  be  applied. 

128.  Integration  of  expressions  containing  Vaoc2  +bx  +  c. 
Every  expression  containing  Va.T2  +  bx  +  c,  but  no  other 
irrationality,  can  be  rationalized  by  a  proper  substitution. 
Two  cases  are  distinguished. 

(a)  When  ax2  -f  bx  +  c  has  real  factors.  We  may  then  write 
the  quadratic  expression  in  the  factored  form 

ax2  -f  bx  +  c  =  a  (x  —  a)  (x  —  /?),  (1) 


250  INTEGRAL   CALCULUS 

in  which   a  and  ft  are  real.     Introduce  a  new  variable   t  by 
means  of  the  formula 


VaX2  +  bx  +  c  =  t(x  —  a) .  (A) 

Square  both  members  of  this  equation  and  replace  the  left 
member  by  means  of  (1).     This  gives 

a(x-a)(x-(3)  =  t2(x-a)2. 

On  canceling  x  —  a  and  solving  for  x  we  obtain  as  the  equa- 
tion of  transformation  .9        0 

x=«tr-aPm  (2) 

Hence  x  (and  therefore  dx)  is  rationally  expressible  in  terms 
of  t,  while  the  radical  reduces  to 


[at2-a(S        1 
_  t2-a 


at(a  —  /?) 
f-a 


(3) 


which  is  also  rational  in  t.  The  substitution  of  these  expres- 
sions in  the  proposed  integrand  gives  a  rational  fraction  which 
may  be  treated  by  the  methods  of  the  preceding  chapter. 

(b)    When  a,  the  coefficient  of  x2,  is  positive. 

Make  the  substitution 

Vase2  +  bx  +  c  =  Va-  x  +  t.  (B) 

By  squaring  both  members  and  solving  for  x  we  obtain 

b  -  2Vat 

while  the  radical  is  expressible  in  the  form 

■Vat2  —  bt-\-^fac  /Kx 

-p >  \») 

2Vat-b 


INTEGRATION   BY   RATIONALIZATION  251 

and  hence  the  integrand  becomes  rational  when  expressed  in 
terms  of  t. 

The  only  case  that  is  not  included  in  (a)  or  (6)  is  that  in 
which  the  factors  of  ax2  +  bx  +  c  are  imaginary  and  the  coeffi- 
cient a  is  negative ;  the  radical  is  then  imaginary  for  all  values 
of  x.  Although  the  integral  can  be  obtained  (in  an  imaginary 
form)  by  either  of  the  preceding  substitutions,  this  case  does 
not  arise  in  practical  applications  of  the  calculus  and  will  not 
be  considered  further. 

Ex.  1.  f dx 

J  x  +  vV-2  +  2  x  -  1 

Formula  (B)  gives 


Vx2  +  2x-l  =  x  +  t, 
whence,  by  solving  for  x,  we  obtain 

,_    '2+l 


2(1  -  0 


_  /2     i     9  j  J.   J 

and  accordingly  dx  =  —  — — -  dt, 

h  y  2(1  -  02 


^  +  Dx-U-fi  +  2<  +  1 


2(1  -  0 

When  these  expressions  are  substituted  in  the  above  integral  it 
reduces  to 


r  (-t*  +  2t+l)dt 


2(1  +  02 

The  work  of  integrating  may  be  facilitated  by  means  of  the  trans- 
formation 1  -f  t  =  2.     The  result,  in  terms  of  x,  is 


h(x  -  VV2  +  2x  -  1)  + 


1  -  x  +  Vx2  +  2  x  -  1 


+  2  log  (1  -  x  +  Vx2  +  2  x  -  1 ). 


252  INTEGRAL   CALCULUS 


Ex.2,   r    Vl  +*</*_. 

^   (1    -  *)  VT  -  X 

By   rationalizing    either    numerator   or    denominator   we   obtain 


Vl  —  x2  as  the  radical  part  of  the  integrand. 


Formula  (A)  gives      Vl  —  x'2  =  7(1  —  x), 


whence  JI±*  =  *,  (1) 

'1  —  x 

or  J-±^  =  <»,  (2) 

1  —  x 


and  hence,  by  differentiation, 

2dx 


(\-xy 


=  2tdt.  (3) 


Add  1  to  both  members  of  (2)  and  combine  the  two  terms  of  the  left 
member.     The  result  is 

(4) 


o 

1   -X 

=  t2+l. 

1) 

ividing  (3)  by  (4), 

we 

have 
rfx 

_2tdt 

Now  multiply  (1)  and  (5)  together  and  integrate.     We  obtain 


f    /l  +  x        dx         r  2  t2  dt 
J^l  -x  '  1  -x~  J  t'2+  1 

=  2jrr^_2tan-iJr±Z 

'1  —  a:  '1  —  a: 


-J 


EXERCISES 


(1  -x)(l  -Vl  -x'2) 

dx 


r dx 

J  V2  x2  -  3  x  +  1  [V2xa-3x-+  1  +  V2(a;  -  1)] 


INTEGRATION   BY   RATIONALIZATION  253 

We  can  rationalize  also  by  means  of  a  trigonometric  substi- 
tution. First  reduce  ax2  -\-bx  +  c  to  the  form  ±t2±k2,  as  in 
Art.  118,  and  then  make  one  of  the  following  transformations: 

In  k2  —  t2  put  t  =  k  sin  0, 
in  t2  —  k2  put  t  =  k  sec  0, 
in  t2  +  k2  put  t  =  k  tan  0. 


Since  V  —  t2  — 1&  is  imaginary,  we  shall  exclude  this  case  from 
consideration. 

The  resulting  trigonometric  functions  can  then  be  integrated 
by  methods  to  be  explained  in  the  next  chapter. 

129.  There  is  one  case  in  which  a  different  transformation 
leads  more  rapidly  to  the  desired  result.  If,  after  reducing 
the  terms  under  the  radical  sign  to  one  of  the  simple  forms 
mentioned  in  the  preceding  paragraph,  the  integrand  can  be 
expressed  as  the  product  of  t  dt  and  a  function  containing  only 
even  powers  of  t,  then  we  may  substitute 


y/±t2±k2  =  Z. 

For  this  gives  t2  =  ±  (z2  ±  k2) 

and  tdt  =  ±zdz, 

and  hence  the  integral  takes  a  rational  form  in  z. 

EXERCISES  ON   CHAPTER   IV 
1      f  (—  ^s  +  4rx)dx 

[Notice  that  Art.  129  is  applicable.] 

2.     f    t(x-a)%-\-\dx  3      C  y/J+ldx 

Jo(x-a)$-(x-a)i'  V7+1  +  2 


254  INTEGRAL   CALCULUS 

4.     f dx  8.     f ,lx 

J  x  +  Vx-1  J  *  +  Vx2  -  1 


5-    j-^4-  9.     fi+^tf* 

*^  («  +  *)*  J  1  +  %x 


6.  f(2- 3  *-*)<**.  10.  f_ 


x3  dx 


x  —  3  x6  +  5  x*  v a'2  —  x2 

7      f  dx  -,,      (y/a*-x*dx 

[Use  trigonometric  substitutions  in  the  following  exercises.] 

</x 


12 


13. 


14. 


*"        x2  ""  (x2  +  a2)s 

f        ^        .  17.     f        rf*        . 

J  (x'2+a2)^  J  (a?-x*)% 


CHAPTER   V 

INTEGRATION   OF   TRIGONOMETRIC   FUNCTIONS 

130.  In  regard  to  the  integration  of  trigonometric  functions, 
it  is  to  be  remarked  in  the  first  place  that  every  rational  trigo- 
nometric function  can  be  rationally  expressed  in  terms  of  sine 
and  cosine. 

It  is  accordingly  evident  that  such  functions  can  be  inte- 
grated by  means  of  the  substitution 

sin  x  —  z. 

After  the  substitution  has  been  effected,  the  integrand  may 
involve  the  irrationality 

Vl  —  2T [    =  COS  X  ]. 

This  can  be  removed  by  rationalization,  as  explained  in  the 
preceding  chapter,  or  the  metnod  of  reduction  may  be  employed. 

The  substitution  cos  x  —  z  will  serve  equally  well. 

It  is  usually  easier,  however,  to  integrate  the  trigonometric 
forms  without  any  such  previous  transformation  to  algebraic 
functions.  The  following  articles  treat  of  the  cases  of  most 
frequent  occurrence. 

131 .  Csec2nx  dx,     f  csc2w  x  dx. 

In  this  case  n  is  supposed  to  be  a  positive  integer. 
If  sec2n#  dx  is  written  in  the  form 

sec2n~-.r  •  setfxdx  =  (1  +  tan2  x)n~ld (tan  a?), 
255 


256  .  INTEGRAL   CALCULUS 

the  first  integral  becomes 

f  (tan2cc  +  l)n-1d(tan  x). 

If  (tan2a?  + 1)"-1  is  expanded  by  the  binomial  formula  and 
integrated  term  by  term,  the  required  result  is  readily 
obtained. 

In  like  manner, 

I  csc2nx  dx  =  f  csc2"~2#  «csc2a;  dx 

=  —  f(cot2^  +  l)n~V/(cot  x). 

This  last  form  can  be  integrated,  as  in  the  preceding  case, 

by  expanding  the  binomial  in  the  integrand. 

The  same  method  will  evidently  apply  to  integrals  of  the 

form 

I  tanm.T  sec2n#  dx,     I  cotTO#  csc2hx  dx, 

in  which  m  is  any  number. 

EXERCISES 

1      f   (Jx  5      f  (1  -cosxydx 

J  cos4z  *  sin4.*: 

r  6      C tjx 

2.  J  CSC4X  dx.  '     J  sin4x.  CQS4X  (CQS4X   _   Sju  V)4 

3.  \secexdx.  7.     i -  [  =  1  tan~8.r  sec4xdx']  . 

J  J  sin3x  cos  x        J 

4      C        ^x  q      rcos2ar  dx 

J  ain^x  cos8*  J     siii°x 


INTEGRATION   OF   TRIGONOMETRIC    FUNCTIONS       257 
132.  fsecwx  tan2n+1sc  dx,     fcscm^  cot2n+1sc  da. 

In  these  integrands  n  is  a  positive  integer,  or  zero,  so  that 
2)i  +  l  is  any  positive  odd  integer,  while  m  is  unrestricted. 
The  first  integral  may  be  written  in  the  form 

I  secm-1aj  tan2na;  •  sec  x  tan  x  dx 

=  I  secm_1a;(sec2#  —  l)'V?(sec  x), 

which  can  be  integrated  after  expanding  (sec2.c  —  l)n  by  the 
binomial  formula. 
Similarly, 

I  cscTO#  cot2n+1#  dx  =  I  cscm_1#  cot2n#  •  esc  x  cot  x  dx 
=  —  I  cscm_1x(csc2a;  —  l)nc?(csc  x). 

EXERCISES 

1.  \  sec2*  tan3x  dx.  5.     I  tan5*  dx. 

2.  f  csc8*  cot5*  dx.  6.     f  sm3-r  dx  [  =  f sec*"8*  tan8*  tfx] . 
•J  J    cos"*  J 

f  sec  a*  ,  r 

Jcot5a*  7/  J  tan  a: 


dx. 


4.     l  :  in  x  cot3*  (/*.  8      i  cot  x  dx. 

EL.    CALC.  17 


258  INTEGRAL   CALCULUS 

133.  CtsiVLnacdx,     Ccotnocdx. 

The  first  integral  can  be  treated  thus  : 

I  tsinnx  dx  =  M  tann~2  •  tan2#  dx 

=  f  tann_2^(sec2ic  —  l)dx 


^""^       'Un-'xdx. 


When  n  is  a  positive  integer,  the  work  of  integration  may- 
be rapidly  carried  out  by  writing  t  for  brevity  in  place  of  tan  x 
and  then  putting  tndx  in  a  different  form  by  means  of  the 
following  process.  First,  divide  tn  by  £2  +  1;  the  quotient  is  a 
polynomial  of  the  form  tn~2  —  tn~~A  +  tn~G  —  •••,  while  the  re- 
mainder R  is  either  ±  1  or  ±  t  according  as  n  is  even  or  odd. 
Then,  since  the  dividend  equals  the  product  of  divisor  and 
quotient  plus  the  remainder,  we  have 

«•=(«•-"  -  r-4  +  r~6 )(t2  + 1)  +  R. 

But  since     (tan2#  +  l)dx  =  sec2x  dx  =  d(tan  x)  =  dt, 
we  have 

Ctsmnx  dx  =  C(tn~2  -  r~*  +  *n~6 )^  +  f  ^  da;- 

For  example, 

f tan8#  dx  =  C{f  -t*  +  t2--  l)dt  +  Cdx, 

and       J  tan7x dx  =  \  (f  —  f  +  t) dt  —  (tan x dx. 


INTEGRATION   OF   TRIGONOMETRIC   FUNCTIONS       259 

The  integral    J  cotn#  dx  can  be  treated  in  a  similar  manner, 
in  case  n  is  a  positive  integer. 
For  any  value  of  n  we  have 

j  cotn#  dx  =  I  cotn-2#  aot2xdx 

=  I  cotn_2#(csc2#  —  l)dx 

=  _COt-1.T_    Ccotn-2xdXt 

n-1       J 

Since  tan  x  and  cot  x  are  reciprocals  of  each  other,  the  above 
method  is  sufficient  to  integrate  any  integral  power  of  tan  x  or 
cot  x. 

Another  method  of  procedure  would  be  to  make  the  substi- 
tution tan  x  =  z,  whence 


j  tann  xdx=  j  — - 


(h 


If  the  exponent  n  is  a  fraction,  say  n  =  -,  the  last  integral 

can  be  rationalized  by  the  substitution  z  =  uq. 

It  is  evident  from  this  that  any  rational  power  of  tangent 
or  cotangent  can  be  integrated. 

EXERCISES 

1.  j  cot*xdx.  3.    (  (tan  x  -  cotx)3dx. 

2.  \  tausaxdx.  4.    (  (tann  x  +  ta,nn-2  x)dx. 

5.    j  tan8  x  dx. 


260  INTEGRAL   CALCULUS 

When  n  is  a  positive  integer  show  that 

^     d.     9n      i        tan2n_1  x      tan2"-3 a;  ,  ,    ,      n„  1A  x 

6.  1  tan2"  x  dx  = h  •••   4-  (—  l)n-1(tan  x  —  x). 

J  2n-  1         2n-3  V        J       K  ' 

r>     Ct.     9„a.i     j        tan2n  x      tan2"-2  a: 

7.  \  tan2"+1  xdx  = 

+  •••  +  (—  l)n_1  (J  tan2  a;  +  log  cos  a;). 
134.     fsium  x  cos**  x  dx. 

(a)  Either  m  or  n  a  positive  odd  integer. 

If  one  of  the  exponents,  for  example  m,  is  a  positive  odd 
integer,  the  given  integral  may  be  written 

j  sin"1-1  x  cosn  x  sin  xdx  =  —  I  (1  —  cos2  x)   2  cosn  x  d  (cos  x). 

Since   m  is  odd,  m  —  1  is   even,  and  therefore   — - —   is  a 

Z 

positive  integer.  Hence  the  binomial  can  be  expanded  into 
a  finite  number  of  terms,  and  thus  the  integration  can  be 
easily  completed. 


E 


x.       (  sin5xVcos  xdx. 


According  to  the  method  just  indicated  this  integral  can  be  re- 
duced to 

—  (  sin4  xVcos  x  d(cos  x)  =  —  \  (1  —  cos2  a;)2(cos  a;)2  c?(cos  x) 

=  —  |  cos  2  X  +  \  COS  *  x  —  T2T  cos  *~  X. 

EXERCISES 

1.  (sin*  xdx.  3.      (- 

2.  j*sin3a:cos4a:rfx.  4>     J  cos2  xj/^ 

_        C    sin8  x  dx 
J  VI  -  cos  a; 


'cos5  X 
sin  x 

dx. 

'     sinf 

'  x  dx 

INTEGRATION   OF   TRIGONOMETRIC    FUNCTIONS       261 

(b)  m  +  n  an  even  negative  integer. 

In  this  case  the  integral  may  be  put  in  the  form 

J^H^.  cosTO+n  x  dx  =  f tanma;  sec-(m+n)  x  dx, 
COSm  X  J 

which   can   be   integrated   by  Art.    131,    since   the   exponent 
—  (in  -+-  n)  of  sec  x  is  an  even  positive  integer. 


fcx.     \ dx. 

COS^  X 

The  integration  is  effected  in  the  following  steps : 


C     y/s\\\xdx  Ct-      h  4      / 

J  — ^= =  \  tan2  x  sec4  x  dx 

Vcos  x  cos4  x     J 

=  \  tan^  x(tan2  x  +  1)  d  (tan  x) 

=  2  tan^2(!  +  \  tan2 a;). 


EXERCISES 


dx 


1.      (^dx  4.      f_ 

J  sin4  ar  ./  sm4x  cos-6  x 

2.  r^_.  5.  f 

J   sin6  a:  * 


Vsin3  x  cos5  x 


3.      f0^*.  6.      p""^**,. 

J  sin8  x  J  cosn+2  x 

(c)  Multiple  angles. 

When  m  and  ?i  are  both  even  positive  integers,  integration 
may  be  effected  by  the  use  of  multiple  angles.  The  trigono- 
metric formulas  used  for  this  purpose  are 

1  —  cos  2  x 


sin-  x  = 


cos-a; 


sin  x  cos  x 


2 
1  4-  cos  2  x 


2 
sin  2  # 


262  INTEGRAL   CALCULUS 

Ex.     (  sin2  x  cos4  x  dx. 


\  sin2  x  cos4  x  dx  =  I  (sin  a:  cos  x)2  cos2  x  dx 


sin2  2x1  +  cos  2  x 


-J 

=  -  f  sin2  2  x  rfx  +  —  f  sin2  2  x  cos  2  x  rf(2  x) 

111- 

=§J-     -dx+ 


1  f  1  —  cos  4  x  ,     ,1  sin3  2  x 
dx  -\ 

2  16     a 


=  j &  x  -  e¥  sin  4  x  +  3X8-  sin8  2  x. 

EXERCISES 

1.  (  cos2  x  sin2  x  </x.  3.      j  sin4  x  cos4  x  tfx. 

2.  (  sin2 x  cos6  x  g?x.  4.      I  (sin4x  —  cos4x)4</x. 

5       C^lldx=  fC1  -cos2x)2^=  f(sec2x-2  +  cos2xyx. 
J  cos2  X  J  COS2  x  J 

(d)  Reduction  formulas.     Integrate   J  sinm#  cosn  x  dx  by  parts, 
taking  u  =  cos*_1a*,   cfa  =  sinm#  cos  x  dx, 


sinm+1a* 


whence  du  =  —  (n  —  1)  cosn  -a;  sin  x  dx,   v  = , 

m+- 1 
and  therefore 

frill**  cos'a!  dx  =     sin"+'a;cos"'"1-c  +  '^4  f sin"*2  cos""2*  dft 
c/  m  + 1  7/1  +  1./ 

In  the  last  term  replace  sin2a-  by  1  —  cos2#  and  separate  the 
integral  into  the  two  terms 

I  sinm#  cosn_2x  dx  —  I  sinma;  cosn#  dx. 


INTEGRATION   OF   TRIGONOMETRIC   FUNCTIONS       263 

Transpose  the  second  integral  and  unite  with  the  similar 
integral  in  the  left  member.      After  dividing   the   resulting 

equation  by we  obtain  the  formula  of  reduction 

m  +  1 

/„     j            sinm+1# cosM_1x  .    n—1   C  -   m         n-i     i 
smm#  cosw#  ax  = 1 I  smm#  cosn  zx  dx 
ra  +  ?i             m  +  n  J 

by  means  of  which  the  exponent  of  the  cosine  factor  may  be 
diminished  or  increased  by  2  according  as  the  integral  in  the 
left  member  or  that  in  the  right  member  is  taken  as  the  given 
integral. 

In  like  manner  a  reduction  formula  may  be  deduced  which 
decreases  or  increases  the  exponent  of  the  sine  factor  by  2. 
The  details  are  left  to  the  student  as  an  exercise.  The 
result  is 

/-  m         n     i           sinw_1x  cos"+1x  .  m—1  C  -  m-2         n    ^ 
smmx  cosnx  ax  = 1 I  smm  lx  cosn#  ax. 
m  4-  n            m  4-  nJ 


The  two  preceding  formulas,  when  solved  for  the  integrals 
in  the  right  members,  and  m  (or  n)  increased  by  2,  become 

/•   m         n    i           sinm+1#  cosn+1./c  .  m-\-n  +  2  C  .   m         n+2   j 
smmaJcosna;aa;= ! ! —  I  smmxcosn+'xax, 
n  +  1                 n  +  l    J 

Csinmx  cosnxdx  =  smW+la?  cosn+1a;  +  m  +  n  +  2  Csinm+2X  G0Snxdx, 
J  m  + 1  m-j-1    J 

Whenever  the  values  of  m  and  n  are  such  that  one  of  the 
three  preceding  cases,  (a),  (6),  (c),  is  applicable,  the  integration 
can  generally  be  performed  more  quickly  by  one  of  those 
methods. 


264  INTEGRAL   CALCULUS 

EXERCISES 

1.    (sin*xdx.  2.    (99&*dx. 

J  J  sin2  x 

[In  Ex.  2  after  one  reduction,  diminishing  the  exponent  of  cos  x  by 
2,  Art.  133  may  be  applied.] 

3      rsin4:r  ,  .      f     dx  _      Cco^xdx 

J  cos  x  J  sin3  2  a;  J    sin4  a; 


135. 


/eta  r doc  r 

a+bnosna    J  a  +  bsmnw    J  a  + 


bcosncc'  J  a  +  bsinnw'  J  a  +  b sin noc+c cos nx 

These  forms  can  be  integrated  by  expressing  them  in  terms  of 
the  half  angle  and  then  in  terms  of  tan  -  -. 


Ex.l.    J 


dx 


5  +  4  cos  x 


By  making  use  of  the  trigonometric  relations 


cos2-  +  sin2-  =  l, 


2  2 


cos  x  =  cos2  - —  sin2  - , 

2  2 


the  denominator  may  be  written  in  the  form 

5^cos2|+sin2|)+  4(cos2|-sin2-V 


which  becomes  sin2  -  +  9  cos2-  on  collecting  the  terms  ;  whence 
2  2 

dx 


1 

sin2! +9 


,,x 

cos2- 

2 


INTEGRATION   OF   TRIGONOMETRIC   FUNCTIONS       265 

Now  divide  numerator  and  denominator  by  cos2  -  and  bear  in  mind 

1  x 

that  =  sec2  -.    This  gives 

„  t  2 

cos2- 

i'(S) 


J 


-  =  ? tan-1  (i  tan- 
tan2- +9       3  Vd        2 


Ex.2,    J- 


t/x 


sin  3  a:  +  1 
Express  the  denominator  in  the  form 

4  sin  —  cos  —  +  (  sin2  — -  +  cos2  — -  J . 

3  a: 
Then,  after  dividing  both  terms  of  the  fraction  by  cos2  — ,  the  given 

integral  becomes         r^  3  x 

sec2  —  dx 
2 


J 


tan2  ^+4  tan—  +  1 


Now  make  the  substitution  tan—  =  t  and  apply  Art.  118. 

It  will  be  observed  from  these  two  problems  that  the  aim  is 
to  put  the  denominator  in  the  form  of  a  homogeneous  quadratic 
expression  in  sine  and  cosine  functions.  Then,  when  both  terms 
of  the  fraction  are  divided  by  the  square  of  the  cosine,  the 
denominator  becomes  quadratic  in  the  tangent  function  while 
the  numerator  can  be  expressed  as  the  differential  of  the  tangent. 

EXERCISES 

1.  ( ^ .  5.     f ^ . 

J  5  +  3  cos  2  x  J  (a  sin  x  +  b  cos  x)2 

2.  (—«* .  6.    f <" 

J  5  —  3  sin  x  J  a2  sin2  x  +  b2  cos2  x 

dx  „      C       dx 


t.  ( — dx     .  7.  f— 

J  1  -  2  sin  2  x  J  1  + 

t.    f * 8.     f— 

J  a  sin  x  -f  b  cos  x  J  1  + 


cos2  a: 


ax 


sin  x  +  2  cos  a: 


266  INTEGRAL   CALCULUS 

136.     I  eax  sin  nx  dx,     I  eax  cos  nx  dx. 

Integrate    |  eax  sin  nx  da;  by  parts,  assuming 

u  =  sin  nx,  and  cfa  =  eax  dx. 
This  gives        ^.      * 

/eoz  sin  nx  c?x  =  -  eax  sin  nx |  eox  cos  nx  dx.  (1) 

Integrate  the  same  expression  again,  assuming  this  time 

u  =  eax,   dv  =  sin  nx  dx. 
Then 

/eax  sin  nxdx  = eax  cos  nx  +  -  (  eax  cos  nx  dx.  (2) 

n  nJ 

Multiply  (1)  by  -  and  (2)  by  -  and  add.     The  integrals  in 

the  right  members  are  eliminated,  and  the  result  is 

J„    .           7        eax(a  sin  nx  —  n  cos  nx) 
eax  sin  nx  dx  =  — * ■ L  • 
a2  +  n2 

By  subtracting  (1)  from  (2),  the  formula 

/,        eax(n  sin  nx  4-  a  cos  nx) 
eax  cos  nx  dx  =  — * — L 
a~  +  iv 

is  obtained. 

EXERCISES  ON   CHAPTER  V 

1.  Derive  the  reduction  formula 

C,ecnx  dx  =  tanssec*-**  +  n-2  recn_2jc  d^ 
J  n  —  1  n  —  \J 

[Integrate  by  parts,  taking  u  =  secM~2:r,  dv  -  sec2a:  dx.~] 

2.  Derive 

f  csc»x  &  =  -  cot*csc;~2*  +  ^?  f  ce-'*  <** 

J  W  —  1  n  —   JV 

J  sin  x  cos  a;  J  cos5  x 


INTEGRATION   OF   TRIGONOMETRIC    FUNCTIONS       267 


5.     f **, 9.     (S™^dx. 

J  cos  x  sin2  x  J      ex 


10.     \  e2x  sin2  x  dx. 


6-  y^-^dx.  10.     f 
J  cos3  x  J 

7-  I  •  11.     i  <?x  sin  2  a:  sin  a:  efa. 

J  (1  -x)Vl  -  a.-2  J 

[Put  a;  =  cos  $] .  [Hint.     2  sin  2  a:  sin  a:  =  cos  x  —  cos  3  a\] 

f  -       a: 
8.     i  e2cos~dx. 

12.  Show  that 

f  sin  ax  sin  fta:  tfa;  =  sin(a  "  &^  -  sin0  +  *>*. 
J  2(a  -  ft)  2(a  +  ft) 

Use  the  trigonometric  formula 

sin  a  sin  ft  =  i  [cos(cc  —  j3)—  cos(ct  +  /?)]. 

13.  Show  that 

r  •                7     7            cos  (a  —  ft)ar      cosfa  +  b)x 
\  sin  aa:  cos  bx  dx  = * *- * — ^    '    • 

J  2(a  -  ft)  2  (a  +  ft) 

14.  Show  that 

f  cos  a*  cos  bxdx  =  sin (g- ft)  a;       sin(q  +  ft)* 
J  2(a-ft)  2(a  +  ft) 

15.  i  sinn  x  cos3a;  da:.  19.     I— — ■ - :~~; 

J  J  sin  x  cos3  x  —  sin3  x  cos  a: 

16.  j"^L_.  20.    f^^-. 

^    '.A 4?  ^  sin4.r  oos4 


sin^a:  cos-*  a: 


sin*  a-  cos*  x 


17.  j*(tan  a:  +  cot  a-)6  dx.  21.    j*  4   ^  <fo. 

18.  ( ««      f  a  sin  a:  +  ft  cos  a: 

■J  (1  +  cos  a:)3  zz'    J  a  Bin  x  +  o  cos~dx- 


ft  COS  X 

[Hint.     Assume 
a  sin  x  +  6  cos  a:  =  .4  (a  sin  x  +  ft  cos  x)  -f-  i?(«  cos  x  —  ft  sin  a:) 
and  determine  J.  and  Z?  by  equating  like  terms.     Treat  Ex.  23  in 
like  manner.] 

23.     (aeX  +  he~Xdx.  24.     fsin(*  +  q><fa. 

^  «e*  +  fte~x  J  sin  (a;  +  ft) 


CHAPTER   VI 


INTEGRATION   AS   A   SUMMATION.     AREAS 

137.  Areas.  The  problem  of  calculating  the  area  bounded 
by  given  straight  or  curved  lines  can  be  solved  by  means  of 
the  Integral  Calculus  provided  that  the  equations  of  the  boun- 
dary curves  are  known  and  satisfy  certain  restrictions. 

Suppose  it  is  required  to  determine  the  area  limited  by  a 
continuous  arc  of  a  curve  whose  equation,  in  rectangular  coor- 
dinates,   is    written 
in  the  form 

y  =/(»),     (i) 

by  the  two  ordinates 
x  —  a  and  x  =  b,  and 
by  the  it*-axis;  that 
is,  the  area  APQB 
(Fig.  59). 

We  proceed  as 
follows.  It  is  as- 
sumed in  the  first 
place,  for  the  sake  of  simplicity,  that/(#)  is  always  increasing 
(or  always  decreasing)  between  x  =  a  and  x  =  b,  so  that  a  vari- 
able point  on  the  arc  PQ  is  continually  rising  (or  falling)  as 
its  abscissa  x  increases.  Suppose,  further,  that  every  ordinate 
between  x  =  a  and  x  =  b  cuts  the  arc  PQ  in  but  one  point.    Let 

2G8 


Fig.  59 


INTEGRATION   AS   A   SUMMATION.     AREAS  269 

the  interval  A  to  B  (Fig.  59)  be  divided  into  n  equal  intervals 
AAV  AlA2'",  An_1  B,  each  of  length  Ax,  so  that 
interval  AB=  b  —  a  =  n  •  Ax. 

At  each  of  the  points  of  division  A,  Alf  A2,  •  ••,  B  erect  ordi- 
nate^ and  suppose  that  these  meet  the  curve  in  the  points 
P,  Px,  P2,  •  ••,  Q.  Through  the  latter  points  draw  lines  PPi, 
P1  R2)  PzR&  ' ' '  Pn-i  Rn  parallel  to  the  x-axis. 

A  series  of  rectangles  PAX,  PXA>,  •  •  •  is  thus  formed,  each  of 
which  lies  entirely  within  the  given  area.  These  will  be  re- 
ferred to  as  the  interior  rectangles.  By  producing  the  lines 
already  drawn,  a  series  of  rectangles  SA1}  SiA2J  •••  is  formed 
which  will  be  called  the  exterior  rectangles.  It  is  clear  that 
the  given  area  will  always  be  greater  than  the  sum  of  the  in- 
terior rectangles  and  always  less  than  the  sum  of.  the  exterior, 
or,  expressed  in  a  formula,  "^ 

PAX  +  iVl2+  •••  +Pn-1B<  Area  APQB  <  SA1  +  S1Ai+  ••• 

■+S.-1-B.  (2) 

The  difference  between  the  sum  of  the  exterior  and  the  sum 
of  the  interior  rectangles  is 
SP1  +  ^fi2+-  +  £„_!  Rn  =  rectangle  Sn_x  T  =  TQ  •  Ax.     (3) 

As  we  suppose  the  curve  to  be  continuous  between  P  and  Q, 
the  line  TQ  is  of  finite  length. 

If  the  number  n  of  equal  parts  into  which  AB  is  divided  is 
increased,  the  first  sum  in  (2)  increases  in  value  and  the 
second  sum  in  (2)  decreases.  Moreover,  as  their  difference 
TQ  •  Ax,  given  in  (3),  approaches  the  limit  zero,  it  follows 
that  the  limit  of  the  sum  of  the  exterior  rectangles  is  equal  to  the 
limit  of  the  sum  of  the  interior  rectangles  when  n  =  cc,  that  is, 
when  Ax  =  0. 


270  INTEGRAL   CALCULUS 

Since  the  required  area  always  has  a  value  intermediate 
between  the  two  sums,  it  follows  that  the  area  is  equal  to  the 
limit  of  either  sum.     So  that,  for  example,  we  have 

^^  =   A^O  tPA>  +  P*A*  +   -   +  Pn~lBl  (4) 

The  second  member  of  this  equation  may  be  expressed  in 
terms  of  the  function  f(x)  which  appears  in  the  equation  (1) 
of  the  given  curve.     For, 

area     PA±  =  AP  •  Ax  =  f(a)Ax, 

since  AP  is  the  ordinate  y  when  x  =  a. 
Similarly, 

area     PiA2  =  AXPX  •  Ax  =  f(a  +  Ax)  •  Ax, 

area    P2A3  =  A2P2  •  Ax  =  f(a  +  2  Ax)  •  Ax, 


area  Pn-YB  =  An_iPn_i  •  Ax  =f(a  +  n  —  1  Ax)  •  Ax. 
If  these  expressions  are  substituted  in  (4),  it  takes  the  form 

area  =  A^o[/(a)+/(a  +  A'T)  +  /(a  +  2Aa;)+  - 


+  f(a  +  n-lAx)-]Ax.  (5) 

As  it  now  stands,  the  formula  just  derived  is  of  little  prac- 
tical value  for  computing  areas.  This  is  due  to  the  fact  that 
there  is  no  general  method  for  calculating  the  sum  of  the  n 
terms  given  in  brackets  in  the  second  member  of  (5). 

Fortunately,  the  value  of  the  limit  of  this  sum  when  n  =  cc 
and  Ax  =  0  can  be  calculated  by  integration  as  we  shall  now 
proceed  to  show. 

138.  Expression  of  area  as  a  definite  integral.  Denote  the 
function  arising  from  the  integration  of  f(x)  by  F(x),  that  is, 


INTEGRATION  AS   A   SUMMATION.     AREAS  271 

let  F(x)  =  ff(x)  dx, 

By  definition  of  the  derivative  of  F(x)  we  have 
lim    F(x  +  Ax)-F(x)_f(  . 


A 


The  quotient  F(x  +  Ax)     F(x)  may  be  written  in  the  form 

f(x)  -J-  <f>,  in  which  <f>  approaches  zero  at  the  same  time  as  Ax, 
otherwise  the  limit  of  the  quotient  when  Ax  =  0  could  not  be 
f(x).     From  this  relation  follows,  on  multiplying  by  Ax, 

F  (x  +  Ax)  -  F  (x)  =f(x)  •  Ax  +  <£  •  Ax.  (6) 

Next,  in  equation  (6)  substitute  for  x  the  successive  values 
a,  a  +  Ax,  a  +  2  Ax,  •  •  •,  a  +  (n  —  1)  Ax. 

We  thus  deduce  the  following  series  of  n  equations,  in  which 
<£i>  02?  ••*  are  used  to  denote  the  different  values  which  <£  may 

take:  F(a  +  Ax)-F(a)=f(a)  -Ax+fa-Ax, 

F(a  +  2  Ax)  —  F(a  +  Ax)=f(a  +  As)  •  Ax  +  <£2  •  Ax, 
F(a  +  3  Aaj)-2P(a  +  2  Ax)=f(a  +  2  Ax)  •  Ax  -J-  <£3  ■  Ax, 


F(a  +  n  —  l  •  Ax)-F(a  +  n-2  •  Ax)=f(a  +  n- 2  ■  Ax)Ax 
+  <£n-i  •  Ax, 


F(a  +  nAx)—F(a  +  n  —  1  •  Ax)  =  /(a  -fn-  1  •  Ax)Ax 
4-  <£*  •  Ax. 

Let  these  n  equations  be  added ;   then  all  but  two  of  the 
terms  in  the  left  member  of  the  sum  cancel  each  other  and  the 


272  INTEGRAL   CALCULUS 

result  may  be  written 


F(b)-F(a)  =  [f(a)+f(a+Ax)  +  ...  +/(a  +  n-l  •  Ax)]&x 
+  W>i  +  02  +  -+0B]Aa>, 
in  which  b  is  written  for  a-\-n  A#,  since  ?i  A#  =  6  —  a. 
Now  let  Aa;  approach  zero.     The  expression 

(<fc  +  <fe-h  ... +0B)Aa> 

vanishes  at  the  limit.  For,  let  <£  denote  the  positive  value  of 
the  numerically  largest  term  of  the  set  <f>u  <f>2,  •  ••,  cf>n;  then  we 
have  evidently 

|(0i  +  <f>2  H +0»)  Aa;|^  (^-|-<I>  ...(w  terms))  A#  =  n<£  •  Aa; 

=  wAa;  •  <£  =  (6  —  a)  •  3>. 

Hence,  from  the  fact  that  ^m  3>  =  0  and  that  b  —  a  is  finite,  it 

Ax=0  ' 

follows  from  Art.  3  that 

Hm(<k  +  4>2+  .••0n)AaJ  =  O; 

and  therefore     2^(6)  -  F(a)  =  J^J[/(a)  +/(a  +  Aa>)  +  ... 


+f(a+n  -  1  •  Aa;)]Aa;.  (7) 

Now  the  right  member  of  this  equation  is  exactly  the  ex- 
pression previously  derived  for  the  area  APQB;  hence, 

area  APQB  =  F(b)  -  F(a).  (8) 

To  compute  the  value  of  the  right  member  of  (8),  first  obtain 
F(x)  by  integrating  f(x)  dx.  Having  determined  F(x),  substi- 
tute the  values  b  and  a  which  x  takes  at  the  extremities  of  the 
arc  bounding  the  given  area  and  then  subtract  the  second  from 
the  first.  This  result  may  conveniently  be  represented  by  the 
symbol 


ff(x)  dx> 


INTEGRATION   AS   A   SUMMATION.     AREAS 


273 


which  indicates  both  the  integration  to  be  performed  and  the 
substitution  of  the  two  limiting  values  a  and  b  for  x.  It  is 
called  the  definite  integral  of  the  function  f(x)  between  the  limits 
a  and  b. 

We  thus  obtain,  as  a  final  formula  for  area, 

area  APQB  =   f  /  (»)  'dx.  (9) 

139.    Generalization  of  the  area  formula.     Instead  of  taking  the 
limit  of  the  sum  of  the  interior  (or  exterior)  rectangles,  a  more 
Y 


Fig.  60 

general  procedure  would  be  to  take  a  series  of  intermediate 
rectangles.  Let  x1  be  any  value  of  x  between  a  and  a  +  Ax,  x2 
any  value  between  a -{-Ax  and  a  -+-  2  Ax,  etc.  Then  /  (xr)  Ax 
would  be  the  area  of  a  rectangle  KLAXA  (Fig.  60)  intermediate 
between  PAX  and  SAX ;  that  is, 

PA.Kf^AxKSA,. 
Likewise  PXA2  <f(x2)Ax  <  S^A2,  etc. 


EL.     CAI.C 


18 


274  INTEGRAL   CALCULUS 

Hence, 

sum  of  interior  rectangles  <  [/(a?i)  +/(^)  +  •••]^aJ 

<  sum  of  exterior  rectangles, 
and  therefore  (cf .  Fig.  59), 

area  APQB  =  ^o  [/(&)  +  f(x2)  +  ...  +  /«)]  Ax.  (10) 

This  result  combined  with  (9)  gives  for  the  definite  integral 
the  more  general  formula : 

fj(x)dx=£l0[f{x1)+f(x2)+  -f(x„)-]Ax.         (11) 

140.  Certain  properties  of  definite  integrals.  From  the  defini- 
tion of  the  definite  integral  |  f(x)dx  as  the  limit  of  a  par- 
ticular  sum,  certain  important  properties  may  be  deduced. 

(a)  Interchanging  the  limits  a  and  b  merely  changes  the  sign 
of  the  definite  integral. 

For,  if  x  starts  at  the  upper  limit  b  and  diminishes  by  the 
addition  of  successive  negative  increments  (—  Ax),  a  change 
of  sign  will  occur  in  formula  (7),  giving 

F(a)-F{b)=£f{x)dx. 

Hence, 

f  af(x)  dx  =  -  Cf{x)  dx. 

(b)  If  c  is  a  number  between  a  and  b,  then 

fbf(x)  dx  =  Cf(x)  dx  +  Cf(x)  dx. 


INTEGRATION  AS   A   SUMMATION.     AREAS 


275 


(c)   Tlie  Mean  Value  Theorem. 

The  area  APQB  (Fig.  61),  which  represents  the  numerical 
value  of  the  definite  integral  may  be  expressed  as  follows. 

Let  an  ordinate  MN  be  drawn 
in  such  a  position  that 

area  PSN  =  area  NRQ. 

If  £  denotes  the  value  of  x  cor- 
responding to  the  point  N,  then 
MN  =  f($),  and 

area  APQB  =  rectangle  ASRB 
=  MN-AB=f{Z){b-a). 

Hence, 

Cft*)**  =  f(gft> -a),     (12) 

%J  a 

in  which  £  is  some  value  of  x  between  a  and  b.     This  result 
is  known  as  the  Mean  Value  Theorem  (compare  Art.  39), 

dx 


is  called  the  mean  ordinate 


f/(*)< 

and  the  ordinate /(£)  =^a    - 

b  —  a 

between  x  =  a  and  x  =  b.     This  is  also  called  the  mean  value 

of  the  function  /(a?)  between  these  limits. 

The  theorem  may  be  expressed  in  words  as  follows : 

The  value  of  the  definite  integral 


£f{x)dx 


is  equal  to  the  product  of  the  difference  between  the  limits  by  the 
value  of  the  function  f(x)  corresponding  to  a  certain  value  x  =  f- 
between  the  limits  of  integration. 

(d)  It  is  frequently  desirable  to  make  a  change  of  variable 
in  the  definite  integral  in  order  to  facilitate  the  work  of  inte- 
gration.    It  is  obvious,  from  the  nature  of  the  definite  integral, 


276  INTEGRAL   CALCULUS 

that  the  limits  of  integration  must  be  changed  so  that  in  the 
new  integral  the  limits  shall  be  the  values  of  the  new  variable 
corresponding  to  those  of  the  old  variable. 


Ex.   Evaluate    \    v  a2  —  x2dx. 


j>~2 


Make  the  change  of  variable  x  —  a  sin  0,  whence  dx  =  a  cos  0  dO, 
and  therefore 


n 

f  °  y/d'-x*  dx  =  a2(  *  cos2  0  d6. 


Here  the  limits  for  the  new  integral  are  determined  by  inspection 

of  the  equation  connecting  x  and  0,  namely,  sin  6  =  -.     It  is  seen  that, 

a 
as  x  varies  from  0  to  a,  sin  6  varies  from  0  to  1.     This  corresponds  to 

a  variation  of  6  between  the  limits  0  and  -.     The  indefinite  integral 
is,  by  Art.  134  (c), 

The  substitution  of  the  limits  gives  the  value  — • 

4 

141.  Maclaurin's  formula.  As  an  application  of  the  mean 
value  theorem  (Art.  140  (c)),  we  derive  Maclaurin's  formula 
with  the  remainder  term. 

Let  s  and  t  be  independent  variables.  Suppose  J[s  — 1\  to- 
gether with  its  first  n  derivatives  with  respect  to  t,  to  be 
continuous  within  the  interval  0  to  tv  Then  we  have  by  inte- 
gration 


£f\s-t)dt  =  -f{s-t) 


=f(s)-f(s-tl). 


On  the  other  hand  if  we  integrate  by  parts,  taking  u  =f'(s  —  0> 
dv  =  dt,  we  obtain 

£lf'(s  -  t)dt  =f'(s  -  t)  •  tT  +  jTV"(«  -t)-tdt 

*/o 


INTEGRATION   AS   A   SUMMATION.     AREAS  277 

Integrate  the  last  term  by  parts,  taking  u  =/"{s  —  t),  do  =  t  dt. 
By   successive   applications   of   this    process   we   deduce   the 
formula 
/(*)_/(*  -  tl)=f'(s  - (,)«.  +/'(•-■  ttfl  +f'"(s  -  trfl  +  - 


+  ~^-r-.   Cf\s-t)t^dt. 
(n  —  1)  \Jo 


(n-l) 
By  the  mean  value  theorem  we  have 

f  V°(*  -  0  tn~l  dt  =fn)(s  -  et^idhf-1  •  tx, 
Jo 

in  which  6  is  a  positive  fraction  and  Qtx  is  the  same  as  £  of 
(12).  Inserting  this  in  the  preceding  equation  and  substituting 
s  =  x,  t-L  —  x  —  a  (hence  s  —  tx  =  a)  we  obtain  as  a  final  form 

f(x)=f(a)  +f'{a)(x-a)  +-^(x  -  af  +  ... 

+  (iS)l/<")(a;  _  6{X  -  a^X  -  °>"- 

If  we  replace  0  by  1  —  0',  the  remainder  term  takes  the  form 
given  on  p.  153,  with  6'  written  in  the  place  of  0. 

142.    Remarks  on  the  area  formula,     (a)  It  is  noticed   that 
the  formula 

Cf(x)dx=  *™  [/(a)  +  /(a  +  Ax)  +  -  +/(o  +  ^=1  .  Az)]A* 

indicates  two  steps,  —  a  summation,  and  a  process  of  passing 
to  a  limit.  The  differential  /(x)  dx  which  appears  under  the 
integral  sign  may  be  regarded  as  representing  the  general 
term/(x)  Ax  of  the  series  to  be  summed,  while  the  process  of 
taking  the  limit  of  this  sum  is  indicated  by  replacing  Ax  with 
the  differential  dx  and  prefixing  the  sign  of  integration. 


278  INTEGRAL   CALCULUS 

The  general  term  f(x)  Ax  represents  the  area  of  an  arbitrary 
rectangle  (of  the  set  of  interior  rectangles)  whose  altitude  is 
the  ordinate  corresponding  to  an  arbitrary  x  and  whose  width 
is  Ax.  This  is  called  an  element  of  area.  The  definite  integral 
may  then  be  thought  of  as  indicating  the  limit  of  the  sum  of 
all  contiguous  elements  of  area  between  x  —  a  and  x  =  b. 

This  notion  of  summation  (followed  by  passing  to  the  limit 
Ax  =  0)  is  a  very  useful  one  in  applying  the  calculus  to  prob- 
lems of  geometry,  mechanics,  and  physics.  In  each  case  an 
application  of  this  notion  consists  in  finding  the  general  ex- 
pression for  an  element  of  the  given  magnitude  (element  of 
area,  element  of  mass,  element  of  moment  of  inertia,  etc.)  and 
then  indicating  the  two  steps  of  summation  and  taking  the 
limit  by  changing  Ax  to  dx  and  prefixing  the  symbol  *  of  the 
definite  integral.  It  must  not  be  forgotten  that  in  every  case 
it  is  necessary  to  prove  that  the  limit  of  the  sum  gives  pre- 
cisely the  desired  result,  f  This  we  have  already  done  in  case 
of  the  area  formula. 

(6)  The  element  of  area/(x)  •  Ax  is  positive  when  the  cor- 
responding rectangle  is  above  the  x-axis,  since  in  that  case  f(x) 
is   positive,  while   Ax  is  positive  if   b>a.     Accordingly,  the 

formula    \   f(x)  dx  gives  a  positive  value  for  an  area  above  the 

x-axis  provided  we  take  b>a. 

Similar  considerations  show  that  the  same  formula  gives  a 
negative  value  for  an  area  below  the  x-axis. 

(c)  If  the  curve  y=f(x)  crosses  the  x-axis  between  the  two 
points  A,  B,  then  the  area  consists  of  a  positive  part  APC, 

*  This  symbol  originated  historically  from  the  initial  of  the  word  sum. 
f  In  some  cases  the  limit  of  the  sum  is  used  as  a  definition  of  the  magnitude 
in  question,  as,  for  example,  in  the  definition  of  the  length  of  arc.     (Art.  151.) 


INTEGRATION   AS  A   SUMMATION.     AREAS 


279 


B 


Fig.  (52 


Q 


represented  by  the  integral    J   f(x)  dx,  and  a  negative  part  CBQ 

represented  by  the  integral    I    f(x)  dx.     The  sum  of  these  two 
integrals,  which  (by  Art.  140  b)  is    Y 
equal  to    I    f(x)  dx,  would  accord- 
ingly give  the  algebraic  sum  of  the    O 
positive  and  the  negative  area, 

((f)  Some    of    the    restrictions 
placed  upon  the  function  f(x)  in  Art.  137  can  be  removed.     In 
the  first  place,  suppose  that  f(x)  is  not  always  increasing  (or 

decreasing)  as  x  increases  from 
a  to  b.  Let  ordinates  be  drawn 
at  the  maximum  and  minimum 
points  of  the  given  arc  PQ  (Fig. 
63).  These  divide  the  required 
area  into  several  parts  A',  A", 
A'"  for  each  of  which  the  ordinates  satisfy  the  original  condi- 
tion of  Art.  137,  hence  we  conclude  that 

area  =  A'  +  A"  +  A"'  =Cf{x)  dx -f  C*f(x)  dx  +  Cf(x)  dx 


=  Cf(x)  dx,  by  Art.  140  (6). 


A  discussion  of  the  methods  to  be  employed  in  case  f(x)  be- 
comes discontinuous,  or  is  not  singly  valued  in  the  assigned 
interval,  is  postponed  to  Art.  143. 

(e)  Since  f(x)=y,  formula  (9)  may  be  written  more  briefly 


area 


APQB=   Cydx. 


(13) 


280 


INTEGRAL   CALCULUS 


X 

B 

y=br 

/ 

/q 

{' 

V=a'    L 

p 

X 

Fig.  04 


(/)  By  exactly  the  same  process  used 
in  deriving  (9),  or  (13),  it  may  be  shown 
that  the  Me&A'PQB'  (Fig.  64)  bounded  by 
the  curve  PQ,  the  ?/-axis,  and  the  two  lines 
y  =  a',  y  =  b'  is  given  by  the  formula 

area  A'PQB'  =  |     x  dy, 

(g)  If  it  is  required  to  find  the  area  bounded  by  several 
arcs  such  as  PQ,  QR,  ES,  etc.  (Fig.  65),  we  may  calculate  by 
formula  (9)  the  simple  areas 
APQB,  BQRC,  etc.,  and  by 
proper  additions  and  sub- 
tractions obtain  the  desired 
area.  Thus  the  area  in  Fig. 
65  would  be  expressed  by 

Xb  /*c  r*c  S*d 

fY{x)  dx  +  J&  f2(x)  dx  -  |   f3(x)  dx  -J    f4(x)  dx. 


1.  Find  the  area  bounded 
by  the  curve  y  =  log  x,  the 
ar-axis,    and    the     ordinates 


—  o    <r  — 


*  The  symbol 


la 


-  Area  APQB  (Fig.  66)  = 

/*8  -i8* 

I    \ogxdx  =  x(logx  —  1) 
=  3(log3-l)-2(log2-l)( 
Fig.  66  =log-^-l. 

indicates  that  the  values  3  and  2  are  to  be  substituted  for 


x  in  the  expression  which  precedes  the  symbol  and   the  second  result  sub- 
tracted from  the  first. 


INTEGRATION   AS   A   SUMMATION.     AREAS  281 

2.  Find  the  area  bounded  by  the  arc  of  the  parabola  y2  =  4px 
measured  from  the  vertex  to  the  point  whose  abscissa  is  a,  the  x-axis 
and  the  ordinate  x  =  a. 

From  the  result  show  that  the  area  of  the  parabola  cut  off  by  a 
line  perpendicular  to  the  axis  of  the  curve  is  two  thirds  the  area  of 
the  rectangle  circumscribing  this  segment. 

Does  this  result  hold  good  for  all  parabolas? 

3.  Find  the  area  between  the  x-axis  and  one  semi-undulation  of 
the  curve  y  =  sin  x. 

4.  Find  the  area  bounded  by  the  semicubical  parabola  y2  =  25  x3 
and  the  line  x  =  3. 

5.  Find  the  area  bounded  by  the  curve  y2  =  4(x-f  5)3  and  the^-axis. 

6.  Find  the  area  bounded  by  the  cubical  parabola  y  =  x3,  the 
y-axis,  and  the  line  y  =  1. 

7.  Find  the  area  bounded  by  the  curve  x  +  y3  =  2  and  the  coordi- 
nate axes. 

8.  Find  the  area  bounded  by  the  parabola  y  =  2  x2  and  the  line 
y  =  2x. 

9.  Find  the  area  bounded  by  the  parabola  y  =  x2  and  the  two  lines 
y  —  x  and  y  =2  x. 

10.  Find  by  integration  the  area  of  the  circle  x2  +  y2  =  r2. 

11.  Find  the  area  between  the  curve  y  =  x(x  —  l)(x  —  3)  and  the 
x-axis. 

12.  Find  the   area  bounded   by  the   coordinate   axes,  the  witch 

8a3  . 

y  =  — — - — -,  and  the  ordinate  x  =  xv     By  increasing  x\  without  limit, 
x2  -f  4  a2 

ftnd  the  area  between  the  curve  and  the  x-axis. 

13.  Find  the  area  of  the  ellipse  —  +  y—  =  1. 

a2     b2 

(*  14.   Find  the  area  included  between  the  hyperbola  xy  =  36  and  the 
-  line  x  -f  y  =  15. 


282  INTEGRAL   CALCULUS 

15.  Find  the  area  bounded  by  the  logarithmic  curve  y  =  ax,  the 
x-axis,  and  the  two  ordinates  x  =  xv  x  =  x2.  Show  that  the  result  is 
proportional  to  the  difference  between  the  ordinates. 

16.  Find  the  area  between  the  curve  y  =  (x2  —  l)(x2  —  2)  and  the 
x-axis. 

17.  Find  the  area  cut  off  from  the  parabola  (x  —  l)2  =  y  —  1  by 
the  line  y  =  x. 

18.  Find  the  area  of  the  oval  in  the  curve  y2  =(x  —  a)(x  —  b)2, 
given  a  <  b. 

19.  Prove  that  the  area  of  the  curve  a2y2  =  x3(2  a  —  x)  is  equal  to 
that  of  a  circle  of  radius  a.  Draw  figures  of  the  two  curves  (center 
of  the  circle  at  the  point  (a,  0))  and  compare. 

20.  Find  the  area  of  the  loop  of  the  curve  y2  =  x4  4-  xB. 

21.  Given  the  curve  of  damped  vibrations  y  =  e~x  sin  x.  Show 
that  the  areas  contained  between  successive  semi-undulations  of  the 
curve,  and  the  positive  x-axis  form  a  geometrical  series  of  alternately 
positive  and  negative  terms. 

Find  the  sum  of  this  infinite  series  and  verify  that  the  same  result 
may  be  obtained  by  integrating  between  the  limits  0  and  go. 

Find  the  total  area  included  between  the  positive  x-axis  and  the 
curve  (changing  the  negative  areas  to  positive). 

22.  Find  the  area  bounded  by  the  hyperbola  xy  =  a2,  the  x-axis, 
and  the  two  ordinates  x  =  a,  x  —  na. 

From  the  result  obtained,  prove  that  the  area  contained  between 
an  infinite  branch  of  the  curve  and  its  asymptote  is  infinite. 

23.  Find  the  area  contained  between  the  curves  ys  =  x  and  x8  =  y. 

24.  Take  the  segment  of  the  equilateral  hyperbola  xy  =  k2,  be- 
tween two  points  P  and  Q.  Show  that  the  area  between  this  arc  and 
the  x-axis  is  the  same  as  that  between  the  same  arc  and  the  y-nxis. 

25.  Find  the  area  bounded  by  the  parabola  Vx  +  Vy  =  Va  and 
the  coordinate  axes. 


INTEGRATION   AS   A   SUMMATION.     AREAS  283 

26.  Find  the  area  between  the  curve  y2(y2  —  2)  =  x  —  1  and  the 
coordinate  axes. 

27.  Find  the  area  common  to  the  two  ellipses 

a2     b*        '    b*     a2 

28.  Find  the  area  enclosed  by  the  curves  y  =  sin  x,  y  —  cos  x  be- 
tween two  consecutive  intersections. 

29.  Find  the  mean  ordinate  of  the  curve  y  =  tan  x  between  the 

limits  x  =  0  and  x  =  -  (see  p.  275). 
4 

30.  Find  the  mean  value  of  the  function  sin  x  between  the  limits 
0  and  -;  also  of  the  function  e^sinar. 

31.  Find  the  area  of  the  loop  of  the  curve 

0        o  a  —  x 

V2  =  x2 . 

*  a  +  x 

143.    Precautions  to  be  observed  in  evaluating  definite  integrals. 

The  method  given  above  for  determining  plane  areas  in  rec- 
tangular coordinates  involves  two  essential  steps  : 

(1)  To  find  the  integral  of  the  given  function  f(x)  ; 

(2)  To  substitute  for  x  the  two  limiting  values  a  and  b,  and 
subtract  the  first  result  from  the  second. 

Erroneous  conclusions  may  be  reached,  however,  by  an  in- 
cautious application  of  this  process.  The  case  requiring  par- 
ticular attention  is  that  in  which  f(x)  becomes  infinite  for 
some  value  of  x  between  a  and  b,  or  at  a  or  b.  When  that 
happens,  a  special  investigation  must  be  made.  The  method 
of  procedure  will  be  brought  out  in  the  following  examples. 

Ex.  1.  Find  the  area  bounded  by  the  curve  y(x  -  l)2=c,  the 
coordinate  axes,  and  the  ordinate  x  =  2. 


284 


INTEGRAL   CALCULUS 


A  direct  application  of  the  formula  gives 


n    cdx 


^]: 


2  c, 


]6 
is  a  sign  of   substitution,  indicating  that  the 

values  b,  a  are  to  be  inserted  for  x  in  the  expression    immediately 
preceding  the  sign,  and  the  second  result  subtracted  from  the  first. 
This  result  is  incorrect.     A  glance  at  the  equation  of  the  curve 

shows  that  f(x)\  = - becomes    infinite   for   x  =  1.     It  is 

L      (x  —  1)'2J 


Fig.  67 


accordingly  necessary  to  find  the  area  OCPA  (Fig.  67)  bounded  by  an 
ordinate  AP  corresponding  to  a  value  x  —  x',  which  is  less  than  1. 
For  this  part  of  the  area/(x)  is  finite  and  positive,  and  formula  (9) 
can  be  immediately  applied,  with  the  result 


area 


OCPA  =  (*     cdx    ,  = c- T  = c—  -c.     0<x'<l. 

JQ      (x_l}2  O-l)J0  X'-I 


If  now  x'  is  made  to  increase  and  approach  1  as  a  limit,  the  value 
of  the  expression  for  the  area  will  increase  without  limit. 

A  like  result  is  obtained  for  the  area  included  between  the  ordi- 
nates  x  =  1  and  x  =  2.     Hence  the  required  area  is  infinite. 

Ex.  2.  Find  the  area  limited  by  the  curve  y8(x2  —  a2)2  =  Sx3,  the 
coordinate  axes,  and  the  ordinate  x  =  3  a. 


INTEGRATION  AS   A   SUMMATION.     AREAS  285 

Since /(x)    = — becomes  infinite  for  x  =  a,  it  is  necessary 

L      O2  -  a2)tJ 

in  the  first  place  to  consider  the  area  OPA  (Fig.  68)  and  determine 


Fig.  68 


what   limit   it  approaches  as  AP  approaches  coincidence  with   the 
ordinate  x  =  a.     Accordingly 

area  OPA  =  £*     2xdx  ^  =  3(x2  -  a2)$T 
(x2  —  a2)f  -*0 


whence 


=  3(>'2-a2)*  +  3a*, 
[area  OP.4]  =  3ai 


lim 


x1  =  a 
In  the  same  manner,  the  area  A'P'QB  has  the  value 

3a     2  xdx         P    2  l 

6  a^  —  3(x  2  —  a2) s, 


£ 


0<x'<a 


a  <  x'  <  3  a. 


(x2  -  a2)* 
As  x'  diminishes  towards  a,  the  area  increases  to  the  limiting  value 

2 

6  aL     Hence,  by  adding  the  two  results,  the  required  area  is  found 
tobe  8al  +  6a*=9ai 

The  same  result  is  found  by  a  direct  application  of  (9),  viz. : 

J-_2^  =  3(x2_a2)r|"  =  9nS) 

•"     (x2-a2)f  Jo 

so  that  in  this  case  an  immediate  use  of  the  area  formula  gives  the 
correct  result. 


286 


INTEGRAL   CALCULUS 


Some  of  the  details  in  such  problems  as  the  two  preceding 
may  be  omitted.  It  is  unnecessary  first  to  put  x  —  x',  a  value 
]ess  than  the  critical  one,  and,  after  integration  and  substitu- 
tion of  limits,  to  let  x'  approach  the  critical  value  as  a  limit. 
For  this  is  clearly  equivalent  to  taking  the  critical  value  at 
once  as  the  upper  limit  for  the  portion  of  the  area  to  the  left 
of  the  infinite  ordinate  (or  as  the  lower  limit  for  the  area  to 
the  right  of  this  ordinate). 

Thus,  in  case  of  an  infinite  ordinate,  the  rule  of  procedure 
becomes : 

Calculate  separately,  by  formula  (9),  the  two  portions  of  area  on 
each  side  of  the  infinite  ordinate  and  add  the  two  results.  If  one 
of  these  portions  is  infinite,  it  is  not  necessary  to  calculate  the 
other ;  the  required  area  is  infinite. 

The  formula  (9)  for  area  has  been  deduced  under  the  as- 
sumption that  the  limits  a  and  b  are  finite.  It  may  happen, 
however,  that  the  curve  y  =f(x)  approaches  the  x-axis  as  an 
asymptote.  It  .might  then  be  required  to  determine  the  strip 
of  area  extending  to  infinity  between  the  curve  and  its  asymp- 
tote. The  method  of  procedure  for  such  a  case  will  be  ex- 
plained in  the  following  example. 

Ex.  3.  Find  the  area  bounded  by  the  curve  y(x2  +  1)  =  1  and 
the  ar-axis. 


INTEGRATION   AS   A   SUMMATION.     AREAS  287 

This  curve  being  symmetrical  with  respect  to  the  y-axis,  it  is 
sufficient  to  calculate  the  area  in  the  first  quadrant.  As  our  formula 
of  integration  does  not  take  account  of  the  case  b  =  co,  we  integrate 
from  0  to  x'  and  in  the  result  cause  x'  to  increase  without  limit. 
This  limit  will  be  defined  to  mean  the  area  between  the  arc  in  the  positive 
quadrant,  its  asymptote,  and  the  y-axis.  It  is  evident  that  these  steps  in 
the  evaluation  amount  to  a  direct  application  of  the  area  formula, 
using  the  limits  0  and  co  .     The  half  area  is,  accordingly, 

V =  tan-]a;       =  tan-1  co  —  tan-1 0. 

Jo    1  +  x2  Jo 

We  are  here  confronted  with  the  difficulty  that  the  anti-tangent  is 
a  many-valued  function  and  there  is  a  question  as  to  which  of  its 
values  should  be  chosen.  It  is  necessary  in  such  a  case  to  go  back 
and  examine  the  limiting  process  just  explained.  The  area  OPQN  is 
equal  to  tan-1a;'  —  tan-*0.  If  x'  approaches  zero,  this  expression 
should  approach  zero;  and  as  x'  increases  continuously  the  area  also 
increases  continuously.  Accordingly,  whatever  value  we  choose  for 
tan_10,  the  limit  of  tan-1  a:'  should  be  the  value  obtained  by  a  continu- 
ous increase  in  this  function  as  x'  increases  without  limit.  The  sim- 
plest value  for  tan-1 0  is  0.     If  tan-  x'  increases  continuously  from  0, 

it  reaches  the  limit  —  when  x'  becomes  infinite.     Hence 
2 

.lim     (tan-V-tan^O)  =  £• 
x'  =  co  v  2 

If  we  choose  tan_10  =  n7r,  7i  any  integer,  then    ,   .       tan-1a;'  =  nT+  — » 

J  x'  =  x  2 

and  the  difference  gives  * ,  as  before. 

Ex.  4.  Find  the  area  bounded  by  the  curve  y(x2  +  a2)2  =  x  and 
the  positive  x-axis. 

Ex.  5.  Find  the  area  bounded  by  the  curve  y  —  tan-1  x,  the  coordi- 
nate axes,  and  the  line  x  —  1. 

In  this  problem  we  have  to  deal  with  a  many-valued  function  of  x. 
In  fact,  to  each  value  of  x  corresponds  an  infinite  number  of  values  of 


288 


INTEGRAL   CALCULUS 


tan-1  a:.     The  problem,  accordingly,  has  an  indefiniteness,  which  must 

be  removed  by  making  some  additional  assumption. 

The  curve  y  =  tan-1  x  consists  of  an  infinite  number  of  branches, 

corresponding  ordinates  of  which   differ  by  integer  multiples  of  tt. 

Each  branch  is  continuous  for  all  finite 
values  of  x  (see  Fig.  70).  It  is  evidently 
necessary  to  select  one  of  these  branches 
for  the  boundary  of  the  proposed  area, 
and  discard  all  the  others.  Suppose,  for 
example,  the  branch  A  B  is  selected.  The 
ordinate   to   this   branch    has   the  value 

it   when   x  is   zero,    and    increases    con- 

_      5  _ 
tinuously   to   it  -\ —  =  - —   as  x  increases 
4        4 

continuously  to  1.  Hence  the  required 
area  is 


Y 

y^" 

A 

~B 

0 

C 

X 

— ^ 

X 

-1 

Fig.  70 


|    tan-1  x  dx  =  \  x  tan-1  x—  \  log  (a;2  -f  1) 


5* -1  log  2. 

4       2s 


EXERCISES 

1.  Find  the  area  bounded  by  the  curve  y2(x  —  1)=  1,  the  asymp- 
tote x  =  1,  and  the  line  x  =  2. 

2.  Find   the  area  bounded  by  the  curve   y3(x  —  l)4  =  1    and   its 
asymptota,  the  x-axis. 

3.  Find  the  area  bounded  by  the  curve  of  Ex.  2,  the  x-axis,  and 
the  ordinate  x  =  2. 

4.  Find  the  area  inclosed  by  the  curve  x2y2  =  a2(y2  —  x'2)  and  its 
asymptote. 

5.  Find  the  area  bounded  by  the  curve  a2x  —  y(x  —  a),  the  rr-axis, 
and  the  asymptote  x  =  a. 


INTEGRATION   AS    A   SUMMATION.     AREAS  289 

x3 


6.    Find  the  area  between  the  cissoid  y1 


2a 


and  its  asymp- 


tote x  =  2  a. 

7.   Find    the    area    between     the    curve    y2(l  —  x2)  =  1   and    its 

asymptotes. 

144.  Calculation  of  area  when  x  and  y  are  expressible  in  terms 
of  a  third  variable.  When  the  rectangular  coordinates  of  any 
point  of  the  boundary  arc  of  the  required  area  are  given  as 

functions  of  a  third  variable  6,  we  may  substitute  in    I    ydx 

the  expressions  for  y  and  dx  in  terms  of  6  and  integrate  be- 
tween the  corresponding  new  limits  for  0  in  accordance  with 
Art.  140(d). 

Area  of  the  cycloid.  This  curve  is  traced  by  a  point  P  in 
the  circumference  of  a  circle  of  radius  r  as  the  circle  rolls  on 
a  straight  line,  without  sliding. 

Y 


O  M 


Fig.  71 


Let  the  point  Pbe  in  contact  with  the  given  line  at  0  when 
the  circle  begins  to  roll.  Suppose  that  an  arbitrary  arc  PQ 
has  rolled  over  the  segment  OQ.  Let  (x,  y)  denote  the  rec- 
tangular coordinates  of  P,  and  let  6  represent,  in  radian  meas- 
ure, the  angle  at  the  center  C  subtending  PQ ;  then, 


OQ  =  arc  PQ  =  rO. 


EL.    CALC.  —  19 


290  INTEGRAL   CALCULUS 

Dropping  a  perpendicular  PR  on  the  line  CQ,  we  have 

PR  =  r  sin  6,   RC  =  r  cos  0. 
Accordingly, 

x=OM=OQ-MQ  =  rO-r  sin  6  =  r  (0  -  sin  0), 

y  =  MP=  QC-  RC=  r  -  r  cos  0  =  r(l  -  cos  0). 

These  are  called  the  two  parametric  equations  of  the  cycloid, 
0  being  a  varying  parameter.  One  complete  arch  of  the  cy- 
cloid is  generated  as  6  varies  from  0  to  2  7r,  that  is,  as  x  varies 
from  0  to  2  ttt.  The  maximum  ordinate  for  this  arc  occurs  at 
x  —  -n-r,  and  the  arc  is  symmetrical  with  respect  to  this  ordinate. 
The  area  inclosed  by  the  arc  OP  A  and  the  cc-axis  is 

ydx=  \      r(l  -  cos  0)  •  r(l  -  cos  0)d6  =  3  ttt3. 
The  area  is  three  times  that  of  the  rolling  circle. 

EXERCISES 

1.  Find   the   area  of   the  ellipse  when  x  and  y  are  expressed  in 
terms  of  the  eccentric  angle,  x  =  a  cos  <£,  y  =  b  sin  <f>. 

What  is  the  meaning  of  the  negative  sign  in  the  result? 

2.  2  2 

2.  Find  the  area  of  the  hypocycloid  x3  +  y*  =  a3  by  expressing  x 
and  y  in  the  form  x  =  a  cos3  0,  y  =  a  sin3  6. 

3.  Find  the  area  of   the  loop  of  the  folium  of  Descartes 

xs  +  ys  —  3  xy  =  0. 
This  area  may  be  calculated  either  by  expressing  x  and  y  in  the 


6*+  1  03+  1 

obtained  by  putting  y  =  Ox  and  solving  for  a:  and  ?/,  or  by  transform- 
ing to  polar  coordinates  and  using  the  polar  formula  for  area,  Art.  145. 


INTEGRATION  AS  A   SUMMATION.     AREAS 


291 


4.  Find   the   area  within  the  curve  y2  =  (1  —  a;2)8  by  assuming 
x  =  cos  0,  y  =  sin3  0. 

5.  Find  the  area  of  (aa:)*+  (by)*  =  (a2  -  b2)*,  the  evolute  of  the 
ellipse.     (See  Fig.  51,  p.  190.)     Express  x  and  y  in  the  form, 

ax  =  (a2  -  b2)  sin8  0,  by  =  (a2  -  b2)  cos3  0. 

145.  Areas  in  polar  coordinates.  Let  PQ  be  an  arc  of  a  curve 
whose  equation  is  given  in  polar  coordinates  (p,  0).  It  is  re- 
quired to  find  the  area  bounded 
by  this  curve  and  the  two  as- 
signed radii  QP  and  OQ. 

Let  A  and  B  be  any  two 
points  of  the  curve  with  coordi- 
nates (p,  0)  and  (p  +  Ap,  0  +  A<9) 
respectively.  Through  ^4  draw 
an  arc  AC  of  a  circle  with  radius 
p  and  center  0.  The  element  of 
area  OAC  is  a  sector  of  a  circle 
of   angle   A0.     The   arc   AC  is,  FlG>  72 

therefore,  p  A0  and  the  sectorial  area  is  |  p2  A0.     The  limit  of 
the  sum  of  all  such  elements  contained  between  OP  and  OQ  is 


!fV« 


(14) 


That  this  is  the  actual  area  sought  remains  to  be  proved  by 
showing  that  the  sum  of  the  elements  of  area  has  the  required 
area  for  its  limit.  This  may  be  done  by  steps  exactly  analo- 
gous to  those  used  in  Art.  137,  which  would  consist  in 
proving  that  the  sum  of  all  interior  sectors,  such  as  OAC,  has 
the  same  limit  as  the  sum  of  all  exterior  sectors,  such  as  ODB. 
The  details  are  left  to  the  student  as  an  exercise. 


292  INTEGRAL   CALCULUS 

EXERCISES 

1.   Find  the  area  of  the  three  loops  of  the  curve  p  =  a  sin  3  0. 
From  the  symmetry  of  the  figure  it  is  seen  that  one  sixth  of  the 
total  area  is  described  as  0  varies  from  0  to  — .     Hence  the  area  is 


6  j*Q  6  ±  a2  sin2  S$  d$=  %  a2 J0  *  (1  -  cos  6  0)dO 


r_a? 
This  is  one  fourth  the  area  of  the  circumscribing  circle. 


2.  Find  the  area  of  the  lemniscate  p2  =  a2  cos  2  0. 

3.  Find  the  area  of  the  circle  p  =  2  r  cos  0. 

4.  Find  the  area  of  the  cardioid  p  =  r(l  —  cos  $). 

5.  Find  the  area  of  the  circle  p  =  10  sin  0. 

6.  Find  the  area  bounded  by  the  hyperbolic  spiral  p$  =  c .  and 
radii  drawn  to  two  arbitrary  points  (pv  #,)  and  (p2,  02).  Show  that 
the  area  is  proportional  to  the  difference  between  the  radii. 

7.  Find  the  area  of  the  four  loops  of  the  curve  p  =  a  sin  2  0. 

8.  Find  the  area  of  the  loop  in  the  spiral  of  Archimedes  p  =  aO 
generated  between  the  limits  —  -  and  +  -  for  6. 

9.  Find  the  area  bounded  by  the  lituus  p26  =  k  and  two  arbitrary 
radii,  making  angles  0\  and  62  with  the  polar  axis. 

10.  Find  the  area  of  one  loop  of  the  curve  p2  =  a2  cos  nO. 

11.  The  radius  vector  of  the  logarithmic  spiral  p  =  e~e  starts  at 
the  angle  0  =  0  and  rotates  positively  about  the  origin  an  infinite 
number  of  times.  Determine  the  area  swept  over  by  the  radius 
vector. 

12.  Find  the  area  of  the  curve  p4  =  sin2  0  cos  0. 

13.  Find  the  area  within  the  curve  p  =  cos2  0. 

14.  Find  the  area  of  the  innermost  loop  of  the  double  spiral  p  =  02. 

146.    Approximate    integration.       The    trapezoidal   rule.      As 

shown  in  Art.  138,  the  numerical  value  of  the  definite  integral 


INTEGRATION   AS   A   SUMMATION.     AREAS 


293 


f  y  dx  is  the  same  as  that  of  the  area  bounded  by  the  curve 
y  z=f(x),  the  z-axis,  and  the  two  ordinates  x  =  a,  x  =  b. 
When  a,  b,  and  the  coefficients  in  f(x)  are  numerically  given, 
the  approximate  value  of  this  area,  and  therefore  of  the  defi- 
nite integral,  can  be  found  by  adding  the  n  terms  of  the  series 
[/(«)  +  f(a  +  Aa0  +  '  *  *  +  f(a  +  n  —  1  •  Ax)]  Ax.  The  close- 
ness of  the  approximation  improves  with  increasing  values  of 
n.  A  much  more  rapid  method  of  approximation  is  now  to  be 
considered. 

Instead  of  forming  rectangles, 
as  in  Fig.  59,  p.  268,  draw  the 
chords  PPX,  PXP2,  •  • .,  Pn_!.Q,  thus 
making  trapezoidal  elements  of 
area,  APPXAX,  AxPxP2£2i  etc. 
Denote  the  ordinates  at  A,  Aly  A2, 


5—p, 


A    Ax  Az 

Fia.  73 


A^B' 


An_ly  Bby  y0,  ylf  y2,  >- 


?/„_!,  yn  respectively.     Also  for  brevity  write  Ax  =  h. 
the  areas  of  the  several  trapezoids  are 

APPXAX  =  1^  +  yx)h, 
AXPXP2A2  =  l(yi  +  y2)h, 


Then 


A>-iPn-iQB  =  i(yn_1  +  yn)h. 

Hence,  by  adding,  we  obtain  for  the  approximate  value  of 
the  definite  integral  the  expression 

*[*¥*  +  *  +  »  +  •••  +  !,„-,]. 

This  is  known  as  the  trapezoidal  formula  for  the  approximate 
value  of  J  ydx  and  this  method  of  computing  its  numerical 
value  is  called  the  trapezoidal  rule. 


294  INTEGRAL   CALCULUS 

147.  Simpson's  rule.  With  three  ordinates.  Instead  of  draw- 
ing the  chords  PP1}  PXP2  pass  a  parabola,  having  its  axis  ver- 
tical, through  the  three  points  P,  Px,  P2  and  determine  the 
area  of  the  double  strip  bounded  by  the  two  ordinates  y0,  y2, 
the  #-axis,  and  the  parabolic  arc. 

The  equation  of  the  parabola  is  of  the  form 

y  =  k  -+-  Ix  -f-  mx2. 

For  convenience  take  the  origin  at  the  foot  of  the  middle 
ordinate  yx.  Then  the  abscissas  of  the  three  ordinates  may 
be  represented  by  —  h,  0,  +  h,  and  the  area  under  the  para- 
bolic arc  is  given  by  the  formula 


x 


\k  +  Ix  +  mx2)dx  =  |  (6  k  +  2  mh2). 


This  result  can  be  expressed  in  a  simple  form  in  terms  of  the 
three  ordinates  2/0,  yl}  y2.     For, 

y0=k  —  lh-\-  mh2, 

Vi  =  k, 

y2  =  k  +  Ih  +  mh2 ; 

therefore,  y0  •+-  y2  =  2  k  +  2  ra/i2, 

hence,  6  ft  +  2  m/i2  =  y0  +  4  ?/j  +  ?/2, 

and,  accordingly, 

parabolic  area  APPXP2A2  —  -  (y0  -f-  4  yx  -f  ?/2).  (IS) 

o 

This  is  Simpson's  parabolic  formula  for  three  ordinates. 

With  n  ordinates.     In  like  manner  the  area  bounded  by  the 
two  ordinates  y2,  y4  and  a  parabolic  arc  through  P2,  P3,  P4  is 

|  (2/2  +  4  2/3  +  2/*),  (16) 


INTEGRATION   AS   A    SUMMATION.     AREAS  295 

and  so  on.     If  the  number  of  ordinates  y0,  yu  •••,  yn  is  odd,  we 
obtain,  by  adding  together  the  expressions  (15),  (16),  etc. 

J[2/o  +  *  Z/i  +  2  2/2  +  4  2/3  +2  1/4  +     •  +  2  2/n-2  +  *  Vn-l  +  2/J- 
o 

This  is  Simpson's  formula  for  the  approximate  value  of   I    ydx. 

148.    The  limit  of  error  in  approximate  integration.     The  ap- 
proximate value  obtained  for   I    f(x)dx  by  means  of  Simpson's 


s 

formula  differs  from  the  true  value  by  an  amount  which  does 

not  exceed* 

_  {Q  —  a)jl\£)rr 
j 


180 


in  which  fIV(£)  is  the  value  of  the  fourth  derivative  of  f(x) 
when  x  is  given  a  certain  value  £  between  a  and  b.  The  limit 
of  error  for  the  trapezoidal  rule  is  * 

(b-a)f"(£W 
12 

Since  £  is  not  definitely  known,  in  applying  the  above 
formulas  to  find  the  limit  of  error  it  is  necessary  to  choose  £ 
so  that  /IV(£)  or  /"(£)  has  its  greatest  value  in  the  interval 
from  a  to  b.  The  result  so  obtained  may  be  considerably 
larger  than  would  be  given  by  the  formula  if  £  were  actually 
known.  In  some  cases  the  result  will  be  so  large  as  to  give  no 
useful  information  in  regard  to  the  closeness  of  our  approxi- 
mation. In  other  cases  it  will  be  small  enough  to  indicate 
that  the  required  degree  of  approximation  has  been  attained. 

For  example,  suppose  it  is  required  to  evaluate 


'login  a 


dx. 


*  See  Markoff,  "Differenzenrechrmnij,"  §  14,  pp.  57,  59. 


296  INTEGRAL   CALCULUS 

Since  f(x)  =  x~l\ogl0x,  we  obtain  by  successive  differentiation 
/^(a;)=a;-6(241og10a;-50*iW"),  M =log10  e==  0.4343,  very  nearly. 
As  we  cannot  readily  determine  by  inspection  the  largest  nu- 
merical value  of  flv(x)  in  the  interval  20  <^  x  <J  30,  we  obtain 

the  next  derivative     .  ,  s  ./rt_.  ,_„«-.. 

/v(a?)  =  ar6(274  Jf- 120  log10x). 

The  first  factor  x~6  is  positive.  The  second  factor  takes  a  nega- 
tive value  for  x  ^20  and  hence /v(<r)  is  negative  in  the  given 
interval.  Therefore,  /IV  (x)  is  a  decreasing  function  for  all  the 
values  of  x  under  consideration.  But  flv(x)  is  positive  for 
x  =  30,  and  accordingly  its  greatest  numerical  value  occurs  for 
x  =  20,  which  is  /Iv(20)  =  0.000003. 

The  limit  of  error  for  Simpson's  formula  is,  therefore, 

_  10(0.000003)  y  =  _  (a0OO0OO2)y. 

If  we  use  3  ordinates,  then  h=5  and  the  error  does  not  exceed 
— 0.0001 +  ;  that  is,  the  error  is  less  than  two  units  in  the  fourth 
place  of  decimals. 

EXERCISES 
In  the  following  problems  use  Simpson's  formula  whenever  an  odd 
number  of  ordinates  is  given.  Determine  the  limit  of  error  and, 
when  possible  by  direct  integration,  the  exact  error.  Also  evaluate 
by  using  the  trapezoidal  rule,  and  compare  the  degree  of  accuracy 
attained  by  the  two  different  methods. 

1.  Evaluate    (    x2  dx  by  the  trapezoidal  rule,  using  5  ordinates ; 
9  ordinates. 

In    the   case   of    9    ordinates,   n  =  8   and    h  =  — ^ —  =  - ,   y0  =  0, 
3/i  =  G)2>!/2  =  l,y3  =  (§)V--,  </8  =  42. 

2.  Prove  that  Simpson's  rule  gives  the  exact  value  of    f  x2dxy 
( h  xs dx,     C\ ax*  +  /?.r2  +  yx  +  8) dx. 


INTEGRATION   AS    A    SUMMATION.     AREAS  297 

3.  Evaluate  (  cos  xdx,  using  3  ordinates  ;  5  ordi nates  ;  7  ordi- 
nates;  9  ordinates.  Notice  the  variation  of  error  with  increasing 
values  of  n. 

4.  Evaluate   (    Vxdx,  using  5  ordinates. 

5.  Evaluate   (    Vl  +  Xs  dx,  using  4  ordinates ;  7  ordinates. 

6.  Evaluate   f    cos  x  dx,  using  7  ordinates. 

Jo 

C 12 

7.  Evaluate   \      log10x  dx,  using  unit  intervals. 

r  to     ,/r 

8.  Evaluate   \     '— ,  using  7  ordinates. 

Jio  loo-     r 

9.  Evaluate   (    Vl  —  xidx,  using  6  ordinates. 

Jo  ° 

10.  Evaluate    (    e~xidx,  using  11  ordinates. 

This  integral  (with  any  upper  limit)  is  called  the  Probability  Inte- 
gral since  it  plays  an  important  role  in  the  theory  of  probabilities. 

IT 

11.  Evaluate  f    Vl  —  3  sin'2 xdx,  using  ?  ordinates. 

r  10 

12.  Evaluate  \     x2dx  by  the  trapezoidal  rule,  using  11  ordinates. 

13.  Calculate  the  value  of  7rfrom  the  formula  —  =  (        (  x    ,  using 

1      J»  1  +  x'2 
11  ordinates. 

Determine  the  error  by  comparison  with  the  known  value  of  7r. 

rl     

14.  Evaluate   i     Vcos6d$,  taking  0  at  intervals  15°,  10°,  0°. 

This,  like  Ex.  11,  is  an  Elliptic  Integral  and  cannot  be  integrated 
by  any  formula  given  in  the  present  volume.  It  occurs  in  the  prob- 
lem of  calculating  friction  in  journals.  (See  "  Engineering  Mathe- 
matics "  by  Prof.  V.  Kakapetoff,  Part  I,  p.  16.     Wiley,  1912.) 


15.    Evaluate   (       °%wx  dx,  using  3  ordinates. 
J20       x  & 


CHAPTER  VII 


GEOMETRICAL  APPLICATIONS 


149.  Volumes  by  single  integration.  The  volumes  of  various 
solids  may  easily  be  calculated  by  a  summation  process  exactly 
similar  to  that  used  in  computing  areas.  The  following  prob- 
lems will  make  the  mode  of  procedure  clear. 

Ex.  1.  A  woodman  fells  a  tree  2  ft.  in  diameter,  cutting  halfway 
through  on  each  side.     The  lower  face  of  each  cut  is  horizontal  and 

the  upper  face  makes  an  angle  of 
60°  with  the  lower.  How  much 
wood  does  he  cut  out? 

The  portion  cut  out  on  one  side 
forms  a  solid  bounded  by  a  cylindri- 
cal   surface   whose    equation    may 


be  taken  in  the  form  x2  + 


1, 


and  by  two  planes  whose  intersec- 
tion may  be  chosen  for  the  y-axis. 
Imagine  this  wedge-shaped  solid 
divided  into  thin  plates  by  means 
of  planes  parallel  to  the  a;2-plane 
and  at  equal  distances  Ay.  The 
volume  of  an  arbitrary  plate  PQRP'Q'R'  is  approximately  equal  to 
the  area  of  the  triangular  face  multiplied  by  the  thickness  Ay. 


Area  PQR  =  \  RP  ■  PQ=  \xz  =^xi 


GEOMETRICAL   APPLICATIONS  299 


since  -  =  tan  60°  =  V3.     The  element  of  volume  is  therefore 
x 


V3  x2    A 


Since  the  figure  is  symmetrical  with  respect  to  the  arz-plane,  it  is  suffi- 
cient to  calculate  the  volume  between  the  limits  0  and  1  for  y  and 
double  the  result. 

The  limit  of  th3  sum  of  all  elements  of  volume  in  the  first  octant 
is 

^j/^=^j-;(i-ys)dy=-L. 

That  this  limit  is  the  volume  to  be  determined  may  be  seen  on 
observing  that  the  element  of  volume  falls  short  of  the  total  amount 
contained  in  the  plate  PQRP'Q'R'  by  the  prismatic  piece  PNP'QMQ'. 
The  sum  of  all  these  neglected  portions,  in  the  first  octant,  is  less 
than  the  volume  of  the  maximum  plate  (having  the  a;z-plane  for  base), 
and  hence  approaches  zero  as  Ay  diminishes. 

Therefore  the  total  volume  of  wood  cut  out  is  cu.  ft. 

V3 

Ex.  2.  Calculate  the  volume  in  Ex.  1,  by  dividing  the  solid  of 
Fig.  74  with  equidistant  planes  parallel  to  the  yz-plane. 

Ex.  3.   Find  the  volume  of  the  ellipsoid 

Imagine  the  solid  divided  into  a  number  of  thin  plates  by  means 
of  planes  perpendicular  to  the  x-axis  and  at  equal  distances  Ax.  Re- 
gard the  volume  of  each  plate  as  approximately  that  of  an  elliptic 
cylinder  of  altitude  Ax,  whose  base  is  the  section  of  the  ellipsoid  by 
one  of  the  cutting  planes.  If  the  equation  of  this  plane  is  x  =  A, 
the  equation  of  the  elliptic  base  of  the  plate  is  (in  y,  z  coordinates) 


300  INTEGRAL   CALCULUS 


**      z*  _         A2 


+  -=1- 
r^2      c2  a2 


X2 
Dividing  by  1  —  — ,  we  obtain 
a2 


The  semiaxes  of  the  ellipse  are 


=  1. 


Since  the  area  of  the  ellipse  is  the  product  of  the  semiaxes  multi- 
plied by  7r  (Ex.  13,  p.  281),  it  follows  that  the  area  of  the  elliptic  base 

is  irbcy  1 -J.     On  replacing  A.  by  x,  the  element  of  volume  may 

be  written 

The  sum  of  all  such  elements  for  values  of  x  varying  by  equal 
increments  Ax  between  0  and  a  differs  from  the  volume  of  the  half 
ellipsoid  by  a  series  of  ring-shaped  portions,  the  total  sum  of  which 
is  less  than  the  volume  of  the  maximum  plate  of  the  figure.  It 
readily  follows  from  this  that  the  total  volume  of  the  ellipsoid  is 


Ca        I        x2\  4 

2  \    irbc    1 )  dx  =  -  rrabc. 

Jo  V         a2/  3 


Ex.  4.    Solve  Ex.  3  by  taking  the  cutting  planes  parallel  to  the  xz- 
plane  and  at  equal  distances  At/. 

Ex.  5.    Solve  Ex.  3  by  taking  the  cutting  planes  parallel  to  the 
x^-plane. 

Ex.  6.    Find  the  volume  of  the  portion  of  the  elliptic  paraboloid 

v2      z2 

'—  H —  =  x  cut  off  by  the  plane  x  —  1. 

a2      b2 


GEOMETRICAL   APPLICATIONS 


301 


Ex.  7:    Find  the  volume  of  the  elliptic  cone  ^  -\ —  =  (x  —  l)2  raeas- 

,  a2      b- 

ured  from  the  ^2-plane  as  base  to  the  vertex  (1,  0,  0). 

Ex.  8.  Find  the  volume  of  a  pyramid  of  altitude  h  and  of  base 
area  A. 

[Hint.  Take  the  base  on  the  ar^-plane,  the  altitude  coinciding 
with  the  2-axis.  Cut  the  solid  into  thin  plates  by  planes  parallel  to 
the  base.] 


e=i. 

b2 


On  the  major  axis  a  plane 
D 


Fig.  75 


Ex.  9.    Given  an  ellipse  — 
a2 

rectangle  A  BCD  is  con- 
structed perpendicular  to 
the  plane  of  the  ellipse. 
Through  any  point  P  of 
the  line  CD  a  plane  is 
constructed  perpendicu- 
lar to  CD.  The  two 
points  ii  and  5  in  which 
the  latter  plane  meets  the 
ellipse  are  joined  to  P 
by  straight  lines.  The 
totality  of  all  lines  so  determined  forms  a  ruled  surface  called  a  conoid. 
Given  A  C  —  p,  find  the  volume  of  the  above  conoid. 

Ex.  10.  A  rectangle  moves  from  a  fixed  point  P  parallel  to  itself, 
one  side  varying  as  the  distance  from  P,  and  the  other  as  the  square 
of  this  distance.  At  the  distance  of  2  ft.,  the  rectangle  becomes  a 
square  of  3  ft.  on  each  side.     What  is  the  volume  generated? 

Ex.  11.  The  center  of  a  square  moves  along  a  diameter  of  a  given 
circle  of  radius  a,  the  plane  of  the  square  being  perpendicular  to  that 
of  the  circle,  and  its  magnitude  varying  in  such  a  way  that  two  oppo- 
site vertices  move  on  the  circumference  of  the  circle.  Find  the  vol- 
ume of  the  solid  generated. 


302 


INTEGRAL   CALCULUS 


Ex.  12.  A  right  circular  cone  having  an  angle  2  0  at  the  vertex  has 
its  vertex  on  the  surface  of  a  sphere  of  radius  a  and  its  axis  passing 
through  the  center  of  the  sphere.  Find  the  volume  of  the  portion  of 
the  sphere  which  is  exterior  to  the  cone. 

Ex.  13.  Find  the  volume  of  the  paraboloid  —  +  ^—  =  z  cut  off  by  the 
plane  z  =  c. 

Ex.  14.  A  banister  cap  is  bounded  by  two  equal  cylinders  of  revo- 
lution of  radius  r  whose  axes  intersect  at  right  angles  in  the  plane  of 
the  base  of  the  cap.     Find  the  volume  of  the  cap. 

150.  Volume  of  solid  of  revolution.  Let  the  plane  area, 
bounded  by  an  arc  PQ  of  a  given  curve  (referred  to  rectangular 

axes)  and  the  ordinates 
pn-\  ^-^L  at  the  extremities  P  and 

Q,  be  revolved  about  the 
x-axis.  It  is  required  to 
find  the  volume  of  the 
solid   so   generated. 

Let  the  figure  APQB 
be  divided  into  n  strips 
of  width  Ax  by  means 
of  the  ordinates  AlPtJ 
A2P2,.-,  A-iPn-v  In 
revolving  about  the 
x-axis,  the  rectangle  APRXAX  generates  a  cylinder  of  altitude 
Ax,  the  area  of  whose  base  is  tt  •  AP  .     Hence 

2 

volume  of  cylinder  =  -k  •  AP  •  Ax. 

The  volume  of  this  cylinder  is  less  than  that  generated  by 
the  strip  APPXAX  by  the  amount  contained  in  the  ring  gen- 
erated by   the   triangular   piece    PRXPV     Imagine    this   ring 


Fig.  76 


GEOMETRICAL  APPLICATIONS 


303 


pushed  in  the  direction  of  the  x-axis  until  it  occupies  the  posi- 
tion of  the  ring  generated  by  CDE.  If  every  other  neglected 
portion  (such  as  is  generated  by  P^P^)  is  treated  in  like 
manner,  it  is  evident  that  the  sum  is  less  than  the  volume 
generated  by  the  strip  A^P^QB,  and  hence  has  zero  for 
limit  as  Ax  approaches  zero.  Therefore  the  sum  of  the  n  cylin- 
ders generated  by  the  interior  rectangles  of  the  plane,  viz. 

tt(.IF  +  A^2+  -+An_1PnJ)Ax, 
has  for  limit  the  volume  required.     But  the  limit  of  this  sum 
is  the  definite  integral    I    wy2dx,  and  hence 

volume  =  7r  I    y2  dx. 

The  volume  generated  by  revolution  about  the  y-axis  is  found 
by  a  like  process  to  be  expressed  by  the  definite  integral 


rj     x?dy, 


in  which  a'  and  b'  are  the  values  of  y  at  the  extremities  of  the 
given  arc. 

When  the  axis  of  revolution  does  not  coincide  with  either  of 
the  coordinate  axes,  a  similar  procedure  will  usually  give  at 
once  the  element  of  volume,    y  Qy 

Examples  1-3  will  illustrate. 

Ex.  Find  the  volume  of  revo- 
lution of  the  segment  of  the 
parabola  y2  =  x  cut  off  by  the 
line  y  =  x,  the  axis  of  revolu- 
tion being  the  given  line. 

Let  OQ  be  the  axis,  and  P 
any  point  of  the  parabolic  arc.     °\  Fig.  77 


304  INTEGRAL   CALCULUS 

If  v  denotes  the  perpendicular  distance  PR  from  P  to  OQ  and  u  the 
length  of  the  line  OR,  then  the  element  of  volume  is 

7T?>2  Am. 

The  formula  of  analytic  geometry  for  the  distance  from  a  point 
to  a  line  gives 

_  y  —  x  _  Vx  —  x 
V~~    V2~~       V2     ' 

in  which  (x,  y)  are  the  coordinates  of  P.  The  second  form  for  v  is 
obtained  by  substituting  for  y  the  expression  given  by  the  equation 
of  the  parabola. 

Since  Am  is  measured  on  a  line  making  an  angle  of  45°  with  the 
.r-axis,  it  follows  that  Am  =  V'2  ■  Ax. 

Hence  the  required  volume  is 

1    [Vx-  xV 


i 


7T 


V2  dx 


0     V     V2     /  30  V2 

EXERCISES 

1.  A  quadrant  of  a  circle  revolves  about  its  chord.  Find  the 
volume  of  the  spindle  so  generated. 

[Hint.  Take  the  equation  of  the  circle  in  the  form  x2  +  y2  =  r2 
and  the  equation  of  the  chord  x  +  y  =  r.] 

2.  Find  the  volume  of  revolution  of  the  segment  of  the  circle 
x2  +  y2  —  r2  cut  off  by  the  line  x  =  a,  this  line  being  the  axis  of 
revolution. 

3.  Find  the  volume  of  the  truncated  cone  obtained  by  revolving 
about  the  ?/-axis  the  segment  of  the  line  3  #  +  #  =  5  between  the 
points  (2,  -  1)  and  (1,  2). 


4.    Find  the  volume  generated  by  the  revolution    of   the   cissoid 

x3 
= about  the  x-axis  from  the  origin  to  the  point  (xv  y^). 

2a  —  x 

What  is  the  limit  of  this  volume  as  xi  approaches  2  a? 


GEOMETRICAL   APPLICATIONS  305 

5.  Find  the  volume  obtained  by  revolving  the  entire  cissoid  about 
its  asymptote,  the  line  x  =  2  a. 

[Hint.  The  element  of  volume  is  tt(2  a  —  x)'2  Ay.  For  the  pur- 
pose of  integration  express  x  and  y  in  terms  of  a  third  variable  t  by 
means  of  the  equations 

z  =  2asin2*,   2/  =  2a—  1 
'   *  cost   J 


6.   Find  the  volume  of  the  oblate  spheroid  obtained  by  revolving 

2    b* 


the  ellipse  — -f  2_  =  1  about  its  minor  axis. 


7.  Find   the  volume  of  the   sphere   obtained   by  revolving   the 

circle  x'2  +  (y  —  k)2  —  r2  about  the  y-axis. 

8.  The  arc  of  the  hyperbola  xy  =  k2,  extending  from  the  vertex 
to  infinity  is  revolved  about  its  asymptote.  Find  the  volume 
generated. 

What  is  the  volume  generated  by  revolving  the  same  arc  about  the 
other  asymptote? 

9.  Find  the  entire  volume  obtained  by  rotating  the  hypocycloid 
xi  4-  y*  =  a*  about  either  axis. 

10.  Find  the  volume  obtained  by  the  revolution  of  that  part  of 
the  parabola  Vx  -f  Vy  =  Va  intercepted  by  the  coordinate  axes  about 
one  of  those  axes. 

11.  Find   the  volume  generated  by  the  revolution  of   the  witch 
8  a* 


'2+4«2 


about  the  a: -axis. 


12.  Find  the  volume  generated  by  the  revolution  of  the  witch 
about  the  #-axis,  taking  the  portion  of  the  curve  from  the  vertex 
(x  =  0)  to  the  point  (xi,  y\). 

What  is  the  limit  of  this  volume  as  the  point  (xv  y{)  moves  toward 
infinity? 

EL.    CALC. 20 


306 


INTEGRAL   CALCULUS 


13.  Find  the  volume  obtained  by  revolving  a  complete  arch  of 
the  cycloid  x  =  a(0  —  sin  6),  y  =  a(l  —  cos  6)  about  the  a>axis. 

Volume  =  Tri       ifdx  =  ira3l      (1  -  cos  0)3  dO. 

14.  Find     the     volume     obtained     by    revolving     the     cardioid 
p  =  a(l  —  cos  6)  about  the  polar  axis. 

Assume  x  =  p  cos  0,   y  =  p  sin  B. 

Then       dx  =  d(p  cos  6)  =  d[a{\  —  cos  0)cos  &] 

=  a  sin  0(  -  1  +  2  cos  6)  dO. 
Hence 

volume  =  it  f  ifdx  =  -  7m3  f  "sin3  0(1  -  cos  0)2(1  -  2  cos  0)  dd. 

151.    Lengths  of  curves.     Rectangular  coordinates.     Let  it  be 

required  to  determine  the  length  of  a  continuous  arc  PQ  of  a 
curve  whose  equation   is  written   in   rectangular   coordinates 

It  is  first  necessary  to  define  what  is  meant  by  the  length 

of  a  curve.  For  this  pur- 
pose,  suppose  a  series  of 
points  Pj,  P2,  '•',  P„_i  taken 
on  the  arc  PQ  (Fig.  78),  and 
imagine  the  lengths  of  the 
chords  PPX,  PXP2,  ---to  have 
been  determined.  The  limit 
of  the  sum  of  these  chords  as 
the  length  of  each  chord  ap- 
proaches zero  will  be  taken, 
in  accordance  with  accepted 
usage,  as  the  definition  of  the  length  of  the  arc  PQ ;  *  that  is, 
arc  PQ  =  Lt  (chord  PP,  +  chord  P,P2  +  ••  •  +  chord  Pn_xQ).  (1) 

*  That  this  limit  is  always  the  same  no  matter  how  the  points  Pi  are  chosen, 
as  long  as  the  curve  lias  a  continuously  turning  tangent,  and  the  distances 


Fig.  78 


GEOMETRICAL   APPLICATIONS  307 

This  definition   is   immediately  convertible  into  a  formula 
suitable  for  direct  application. 

For,  let  the  points  Px,  P.,,  •  ••  be  so  chosen  that 

PB1  =  P1R2  =    ..  =  Ax, 

the  lines  PBi,  etc.,  being  drawn  parallel  to  the  ic-axis. 

Denote  by  Ay  the  increment  RxPx  of  y.     Then  the  length 
of  the  chord  PPX  is 


V(A*)»  +  (Ayy  =  yjl+  ( '^Y  A*  =  X/l  +  ( |^)  V         (2) 


\AxJ  *         \&yj 

Now  — ^  is  the  slope  of  PPV     It  is,  therefore,  equal  to  the 

l\X 

slope  of  that  tangent  to  the  arc  PPX  which  is  parallel  to  the 
chord.  If  (x1}  i/i)  denote  the  coordinates  of  the  point  of  con- 
tact of  this  tangent  line,  then  we  have 

Ay  _  dyl 
Ax      dxx 

Hence  the  length  of  chord  PPX  may  be  expressed  in  the  form 
/(x^Ax,  in  which 

'w-V1*®1-  (3> 

Similarly 

P,P2  =f(x2)  Ax,    P2PS  =f(x3)  Ax,  •  •  • , 

in  which  x2  is  the  abscissa  of  a  certain  point  on  the  arc  PXP2 , 
and  so  for  xs,  •••.  When  these  expressions  are  substituted  in 
(1),  it  becomes 

arC  PQ  =  Ax™  0 ^"W  +/&)  +  • ' '  +/(*-)]  **• 

Pi-lPi  are  all  made  to  tend  towards  zero,  admits  of  rigorous  proof.  The 
proof  is,  however,  unsuitable  for  an  elementary  textbook.  (See  Rouche  et 
Comberousse,  "  Traite  de  geome'trie,"  Part  I,  p.  189,  Paris,  1891). 


308  INTEGRAL   CALCULUS 


But,  by  (11),  p.  274,  this  limit  is    J   f(x)dx.    -Substituting  for 
f(x)  from  (3),  we  obtain  the  formula 


-"e-rv+ffi)'**  w 


day 

in  which  a  and  b  are  the  abscissas  of  P  and  (^respectively. 
Taking  for  PPX  the  second  form  in  (2),  namely, 


vwiw 


we  deduce  in  like  manner 


-^-xv+sy* 


\d>j; 
in  which  a'  and  bf  are  the  ordinates  of  P  and  Q. 

EXERCISES 

1.   Find  the  length  of  arc  of  the  parabola  y2  =  4  px  measured  from 
the  vertex  to  one  extremity  of  the  latus  rectum. 

In  this  case  ^=-W", 

dx      y  x 


hence  length  of  arc  =  f  P\il  +  £  dx  =  ("    x+p     dx. 

J0    "        x  >   Vx*+px 

2.  Find  the  length  of   arc  of   the   semicubical  parabola  ay2  —  xz 
from  the  origin  to  the  point  whose  abscissa  is  -• 

3.  Find   the   length  of   arc  of  the  curve  y  =  log  cos  x,  measured 
from  the  origin  to  the  point  whose  abscissa  is  —  • 

2  2  2 

4.  Find  the  entire  length  of  the  hypocycloid  x*  +  y^  =  a5. 


GEOMETRICAL   APPLICATIONS  309 


6.    Find  the  length  of  arc  of  the  catenary  y  =  -  (ea  +  e  a)  from 


the  point  (0,  a)  to  the  point  whose  abscissa  is  a. 

6    '  2x 


x3       1 
7.    Find  the  length  of  arc  of  the  curve  y  =  —  +  — -  between  the 


limits  x  =  1  and  x  =  2. 

8.  Find  the  length  of  the  logarithmic  curve  y  =  log  x  from  a:  =  1 
to  x  =  V3. 

9.  Find  the  length  of  arc  of  the  evolute  of  the  ellipse 

(ax)s  +  (byfs  =  (a*-b2)l 
10.   Find  the  length  of  arc  of  the  curve  y  =  a  log  (a2  —  x2)  from 
x  =  0  to  x  =  - . 

152.  Lengths  of  curves.  Polar  coordinates.  The  polar 
formulas  for  length  of  arc  may  be  derived  from  those  of  the 
previous  article  by  transformation  from  rectangular  to  polar 
coordinates. 

Since  x  =  p  cos  0,  y  =  p  sin  6,  we  obtain  by  differentiating 
with  respect  to  6 

dx  =  (dp cos  0-p  sin  dXie,   dy  =  (d?  sin  6  +  p  cos  $)  cl0, 

hence 


Vi+©,*-va^v-v@y+7* 


Therefore  the  length  of  arc  is 

arcPe  =  £^  +  (|y<W,  (5) 

the  limits  of  integration  being  the  values  of  6  at  P  and  Q. 


310  INTEGRAL   CALCULUS 

If  p  instead  of  0  is  taken  as  the  independent  variable,  we 
deduce  in  like  manner 

arcPQ=jf;Vl+(^)V 

the  limits  being  the  values  of  p  at  P  and  Q. 

EXERCISES 

0 

1.  Find  the  length  of  arc  of  the  logarithmic  spiral  p  =  ea  between 

the  two  points  (pv  0j)  and  (p2,  02),  and  show  that  it  is  proportional 
to  the  difference  of  the  two  radii  p\  and  p2. 

2.  Find  the  length  of  arc  of  the  circle  p  =  2  a  sin  0. 

3.  Find  the  entire  length  of  the  cardioid  p  =  a(\  —  cos  6). 

Q 

4.  Find  the  length  of  the  parabola  p  =  a  sec2  -  between  the  points 
(Pv  #])  and  (p2,  (92). 

5.  Find  the  length  of  the  spiral  of  Archimedes  p  =  aO  between 
two  arbitrary  points. 

6.  Find  the  length  of  arc  of  the  spiral  p  =  0'2  measured  from 
0  =  0  to  6  =  7t. 

7.  Find  the  entire  length  of  the  curve  p  =  cos'2  0. 

A 

8.  Find  the  entire  length  of  the  curve  p  =  a  sin3r- 

o 

3.  Find  the  length  of  arc  of  the  cissoid  p  =  2  a  tan  0  sin  6  between 

the  limits  0  and  —  • 
4 

[Hint.     For  the  purpose  of  integration,  express  the  integrand  in 

terms  of  sec  0  as  the  independent  variable.] 

153.  Measurement  of  arcs  by  the  aid  of  parametric  representa- 
tion. Suppose  the  rectangular  coordinates  of  a  point  on  a 
given  curve  are  expressed  in  terms  of  a  third  variable  t.     Then, 


GEOMETRICAL   APPLICATIONS  311 


since  in  rectangular  coordinates  —  =-%/(  — J  +(  —  )    (Art.  41), 
we  have 


in  which  s  =  arc  PQ,  and  tlt  t2  are  the  values  of  t  corresponding 
to  the  points  P  and  Q.  In  like  manner,  if  the  polar  coordi- 
dates  (p,  0)  are  expressed  in  terms  of  t,  the  formula  for  length 
of  arc  is 


-JWaW'Sr' 


dt 


since 


iwijH'f)' 


EXERCISES 

1.  Find  the  length  of  a  complete  arch  of  the  cycloid 

x  =  a(t  —  sin  t),   y  =  a(l  —  cos  /). 

2.  Find  the  length  of  the  epicycloid 

x  =  a(m  cos  t  —  cos  mt),   y  =  a(m  sin  t  —  sin  m/) 

from  t  =  0tot=     ' 


m  —  1 

3.  Find  the  length  of  arc  of  the  hypocycloid  x*  -\-  y^  =  a?  by  ex- 
pressing x  and  y  in  the  form  x  —  a  sin3  t,  y  —  a  cos3  t. 

4.  Find  the  length  of  the  involute  of  the  circle 

x  =  a(cos  t  +  t  sin  <),    y  =  a(sin  t  -  t  cos  f) 
from  <  =  0  to  t  =  tv 

5.  Find  the  length  of  arc  of  the  curve  x*  —  yt  =  a%  from  (a,  0) 
to  (xv  yj  by  assuming  x  =  a  sec3  t,  y  =  a  tan3  /. 


0 


312  INTEGRAL   CALCULUS 

6.  Find  the  length  of  arc  of  the  curve  x  =  e*  sin  t,  y  =  el  cos  t  from 
t  =  0  to  t  =  t\. 

7.  Find  the  length  of  arc  of  the  curve  x  =  a  +  t2,  y  =  b  4  t3,  meas- 
ured from  the  point  t  =  0  to  the  point  t  =  tv 

154.    Area  of  surface  of  revolution.     Let  AQ  be  a  continuous 
arc  of  a  curve  whose  equation  is  expressed  in  rectangular  coordi- 
nates x  and  y.     It  is  required  to 
determine  a  formula  for  the  area 
of  the  surface  generated  by  revolv- 
ing the  arc  AQ  about  the  a>axis. 
It  has  been  shown   in  Art.   44, 
if      p.  81,  that  if  S  denotes  the  area 
of    the    surface  generated  by   the 
FlG-  79  rotation  of  AP  (P  being  a  variable 

point  with  coordinates  (x,  y)),  then  A#  satisfies  the  conditions 

of  inequality 

2  7t  y  As  <  AS  <  2  7r(y  +  Ay)  As.  (6) 

Let  the  arc  AQ  be  divided  into  n  equal  parts  of  length  A.s. 
For  each  segment  of  arc  there  will  be  a  set  of  conditions  such 
as  (6),  the  values  of  y,  Ay,  AS  being  in  general  different  for 
the  different  segments.     Let  the  n  sets  of   inequalities  thus 

obtained  be  added.     In  what  follows,  the  symbol  ^,  will  be 
used  as  an  abbreviation  of  the  expression,  "The  sum  of  the 

n   terms   of   the  form."     Since  2j  A>S  =  S   (in  which  S  now 
denotes  the  entire  surface  generated  by  arc  AQ),  we  have 

27r^yAs<S<2Tr^(y  +  Ay)As.  (7) 

Now  let  As  (and  hence  Ay)  approach  zero.  The  first  mem- 
ber of  (7)  becomes  2-k  |  yds,  which  changes  to 


GEOMETRICAL   APPLICATIONS  313 


on  making  x,  or  y,  the  independent  variable.     The  limit  of  the 
last  member  of  (7)  may  be  written 


lim 

As. 


™o  X  ty  As + Ay  As~i  =fy  ds + limS  A^ As- 

The  last  term  is  zero.     For,  let  8  represent  the  maximum 
value  of  Ay  in  any  of  the  terms  of  2  Ay  As.     Then  follows 

]T  Ay  As  ^  8  ]P  As  =  8  •  arc  AQ, 

and  since  8  approaches  zero,  we  conclude  that  lim  Va^As  =  0. 

Hence  lim  ^?y  As  =  lim  ^P  (y  -f  Ay)  As, 

and  therefore 


In  like  manner  the  area  of  the  surface  obtained  by  revolving 
arc  AQ  about  the  y-axis  is 


EXERCISES 

1.    Find   the  surface  of  the  catenoid  obtained  by  revolving  the 

X  _x 

catenary  y  =  ~  (ea  +e    a)  about  the  y-axis,  from  x  —  0  to  x  =  a. 

,  XX 

Since  -^-  —  \{ea-e    a), 

ax 

it  follows  that 


1  +  (dyy_(e*+e~°)\ 
+  \dx)   '  4 


314  INTEGRAL   CALCULUS 


hence,  by  using  the  first  formula  of  (8),  the  required  surface  has  the 
area 

[ea  +  e    a)dx. 


rjo  *(< 


2.  Find  the  surface  obtained  by  revolving  about  the  y-axis  the 
quarter  of  the  circle  x2  -\  y2  +  2x  +  2y  +  l  =  0  contained  between 
the  points  where  it  touches  the  coordinate  axes. 

3.  Find  the  surface  generated  by  revolving  the  parabola  y2  =  4px 
about  the  £-axis  from  the  origin  to  the  point  (p,  2  p). 

4.  Find  the  surface  generated  by  the  revolution  about  the  ?/-axis  of 
the  same  arc  as  in  Ex.  3. 

5.  Find   the   surface  generated    by  the   revolution  of  the  ellipse 

a2  +  b2  ~    ' 

(a)  about  its  major  axis  (the  prolate  spheroid)  ; 

(b)  about  its  minor  axis  (the  oblate  spheroid). 

6.  Find  the  surface  generated  by  the  revolution  of  the  cardioid 
p  =  a(l  +  cos  0)  about  the  polar  axis. 

Regarding  the  figure  as  referred  in  the  first  place  to  rectangular 
axes  such  that  x  —  p  cos  0,  y  =  p  sin  0  we  have 


surface  =  2  tt  jj y  ds  =  2  tt  f  ""  p  sin  B^p2  +(^Yd0> 


ds  =^p2  +  (& YdO  by  Art.  45. 


7.  Find  the  surface  of  the  cone  obtained  by  revolving  that  por- 
tion of  the  line  -  +  £  =  1  which  is  intercepted  by  the  coordinate  axes, 

a      b 

(«)  about  the  x-axis ;  ((S)  about  the  y-axis. 

8.  Find  the  surface  of  the  sphere  obtained  by  revolving  the  circle 
p  =  2  a  cos  0  about  the  polar  axis.     [Cf.  Ex.  6.] 


GEOMETRICAL   APPLICATIONS 


315 


9.   Find  the  surface  generated  by  the  revolution  of  a  complete 
arch  of  the  cycloid  x  —  a(0  —  sin  0),  y  =  a(\  —  cos  0)  about  the  z-axis. 

10.  Find  the  surface  of  the  ring  generated  by  revolving  the 
circle  x2  +  (y  —  k)2  =  a2,  k>a,  about  the  x-axis.  Also  find  the  vol- 
ume of  this  ring. 

11.  Find  the  surface  generated  by  the  rotation  of   the  involute 

of  the  circle 

x  =  a  (cos  t  +  t  sin  /),   y  =  a  (sin  t  —  t  cos  t) 

about  the  x-axis  from  t  —  0  to  t  =  tv 

155.   Various  geometrical  problems  leading  to  integration. 

Ex.  1.  A  string  AB  of  length  a  has  a  weight  attached  at  B.  The 
other  extremity  A  moves  along  a  straight  line  OX,  drawing  the  weight 


in  a  rough  horizontal  plane  XOY.     The  path  traced  by  the  point  B 
is  called  the  tractrix.     What  is  its  equation  ? 

Let  OF  be  the  initial  position  of  the  string  and  AB  any  intermedi- 
ate position.     Since  at  every  instant  the  force  is  exerted  on  the  weight 


316 


INTEGRAL   CALCULUS 


B  in  the  direction  of  the  string  BA,  the  motion  of  the  point  must  be 
in  the  same  direction ;  that  is,  the  direction  of  the  tractrix  at  B  is 
the  same  as  that  of  the  line  BA  and  hence  BA  is  tangent  to  the  curve. 
The  expression  for  the  tangent  length  is  (Art.  48,  p.  86) 


dy 

dx 


=wir 


+  l  =  a. 


dx 


Solving  for  • — ,  we  obtain 
dy 


dx 


Va- 


Integrating  with  respect  to  y  gives 
Vo2 


-J- 


y 


dy  =  Va*  -  f  -  a  log  «  +  Va2  -  y*  +  c 

y 


The  constant  of  integration  is  determined  by  the  assumption  that 
(0,  a)  is  the  starting  point  of  the  curve.  Substituting  these  coordi- 
nates in  the  above  equation,  we  find  C  =  0. 

Ex.  2.  The  equiangular  spi- 
ral is  a  curve  so  constructed 
that  the  angle  between  the  ra- 
dius vector  to  any  point  and 
the  tangent  at  the  same  point 
is  constant.    Find  its  equation. 

Ex.  3.  Determine  the  curve 
having  the  property  that  the 
line  drawn  from  the  foot  of 
any  ordinate  of  the  curve  per- 
pendicular to  the  correspond- 
ing tangent  is  of  constant 
length  a. 

If    the     angle    which     the 


Fig.  81 


GEOMETRICAL   APPLICATIONS  317 

tangent  makes  with  the  a>axis  is  denoted  by  <f>,  it  is  at  once  evident 
(Fig.  81)  that 

a  .  1  1 

-  =  cos  </> 


VI  +  tan**     A/l+(^)! 


From  this  follows 


x  -  i0g  (?y  +  Vf  -  a2)  +  C. 


When  the  tangent  is  parallel  to  the  x-axis,  the  ordinate  itself  is  the 
perpendicular  a.  If  this  ordinate  is  chosen  for  the  y-axis,  the  point 
(0,  a)  is  a  point  of  the  curve,  and  hence 

C  =  -  log  a. 

The  equation  can  accordingly  be  written 


v  +  ^v  ~a'2= c«.  (i) 

a 
From  this  follows,  by  taking  the  reciprocal  of  both  members, 


y  +  Vyt-  a2 
whence,  on  rationalizing  the  denominator, 


y  -  Vy2  _  qa  _  ^ 

a 

o 

Adding  (1)  and  (2)  and  dividing  by  -,  we  obtain 


(2) 


y  =  |(c«  +  «  «), 

which  is  the  equation  of  the  catenary. 

Ex.  4.    Find  the  equation  of  the  curve  for  which  the  polar  subnor- 
mal is  proportional  to  (is  a  times)  the  sine  of  the  vectorial  angle. 


318  INTEGRAL   CALCULUS 

Ex.  5.  Find  the  equation,  in  rectangular  coordinates,  of  the  curve 
having  the  property  that  the  subnormal  for  any  point  of  the  curve 
is  proportional  to  the  abscissa. 

Ex.  6.  Find  the  equation  in  polar  coordinates  of  the  curve  for 
which  the  angle  between  the  radius  vector  and  the  tangent  is  n  times 
the  vectorial  angle.     What  is  the  curve  when  n  =  1  ?     When  n  =  |  ? 

Ex.  7.  Find  the  rectangular  equation  of  the  curve  for  which  the 
slope  of  the  tangent  varies  as  the  ordinate  of  the  point  of  contact. 

Ex.  8.  Find  the  equation  of  the  curve  for  which  the  polar  sub- 
tangent  is  proportional  to  the  length  of  the  radius  vector. 

Ex.  9.  Find  the  volume  generated  by  the  revolution  of  the  trac- 
trix  (see  Ex.  1)  about  the  positive  x-axis. 

Ex.  10.  Find  the  area  of  the  surface  of  the  revolution  described 
in  Ex.  9. 

Ex.  11.  Find  the  length  of  the  tractrix  from  the  cusp  (the  point 
(0,  a))  to  the  point  (an,  yi). 

Ex.  12.  Derive  the  following  formulas  for  the  length  of  arc  s  of  a 
twisted  curve,  in  space  of  three  dimensions,  limited  by  the  points 
(#i,  yi,  zi),  (#2,  yi-,  Z2),  the  coordinates  being  rectangular: 

-i:v1+(gr+(i)a-£v1+(i)v(i)> 

Ex.  13.    Using  the  formula  of  Ex.  12,  find  the  length  of  the  helix 
x  =  a  cos  t,   y  =  a  sin  t,   z  =  bt, 
in  which  a  and  b  are  constants,  and  t  is  a  variable  parameter. 

Ex.  14.  A  plate  of  steel  is  \  inch  thick  and  has  the  form  of  a  right 
segment  of  a  parabola.  It  weighs  490  lb.  per  cubic  foot.  Find  the 
total  weight  of  a  plate  30  in.  broad  and  16  in.  long. 

Take  the  equation  of  the  parabola  in  the  form  y2  =  ±px.  Since 
y=  15  when  x  —  16,  we  may  find  the  value  of  p  by  substituting  these 


GEOMETRICAL   APPLICATIONS 


319 


coordinates  in  the  assumed  equation,  namely,  4/> 
the  parabolic  plate  is  therefore 


*tf.     The  area  of 


/*16  1 

2  \     l£  x^  dx  sq.  in. 


The  volume  and  hence  the  weight  are  now 
readily  obtainable. 

Ex.  15.  A  plate  of  wrought  iron  of  heavi- 
ness 480  lb.  per  cubic  foot  is  \  in.  thick  and 
is  bounded  by  three  straight  edges  at  right 
angles  to  each  other,  as  shown  in  the  figure, 
while  the  curved  boundary  is  a  hyperbola  Fig.  82 

with  the  equation  (x  +  5)  y  =  40,  the  base  of  the  figure  being  on  the 
ar-axis.      Calculate  the  weight. 

Ex.  16.  A  metal  plate,  in  the  form  of  an 
equilateral  triangle,  is  \  in.  thick  and  has  an 
altitude  of  4  in.  Any  very  narrow  vertical 
strip,  as  AB,  of  length  2y  and  width  Ax,  is 
of  nearly  uniform  density-  The  density  varies 
from  one  strip  to  another  in  such  a  way  that  Fig.  83 

the  weight  y  per  cubic  inch  is  determined  by  the  condition 


=  0.26     1  + 


100  \ 
9  +  xV' 


Find  the  weight  of  the  plate. 

[Hint.     Calculate  the  weight  of  the  strip  AB,  then  take  the  limit 
of  the  sum  of  all   such  strips  con-      T 
tained  in  the  figure.]  r — 

Ex.  17.  A  trapezoidal  plate  ABCD 
is  |  in.  thick.  The  weight  y  per  cubic 
inch  is  constant  along  any  vertical 
line,  but  varies  with  x  according  to 
the  law 

y  =  0.05  x1  oz.  per  cubic  inch. 


320  INTEGRAL   CALCULUS 

The  first  strip  DA  is  4  in.  from  the  origin.  What  altitude  h  must 
be  adopted  for  the  trapezoid  in  order  that  the  total  weight  of  the 
plate  may  be  just  three  ounces? 

Ex.  18.  The  frustum  of  a  paraboloid  of  revolution  has  vertical  par- 
allel bases  five  inches  apart.  The  equation  of  the  meridian  curve,  with 
the  inch  as  the  linear  unit,  is  y  =  Vx.  The  heaviness  y  is  constant 
over  a  vertical  plane  section,  but  varies  with  x  according  to  the  law 


y  =  0.06  VlOO  —  x2  lb.  per  cubic  inch.      Find  the  total  weight  from 
x  =  4  to  x  =  9. 


CHAPTER  VIII 

SUCCESSIVE   INTEGRATION 

156.    Functions  of  a  single  variable.     Thus  far  we  have  con- 
sidered the  problem  of  finding  the  function  y  of  x  when  -&■ 

dx 

only  is  given.      It  is  now  proposed  to  find  y  when  its  nth. 

d"v 
derivative  — &  is  given. 
dxn 

The  mode  of  procedure  is  evident.     First  find  the  function 

— '"  which  has  — -  for  its  derivative.     Then,  by  integrating 
dxn~l  dxn 

dn~2v 

the  result,  determine   -„,  and  so  on  until  after  n  successive 

dxn~2 

integrations  the  required  result  is  found.      As   an   arbitrary 

constant  should  be  added  after  each  integration  in  order  to 

obtain  the  most  general  solution,  the  function  y  will  contain 

n  arbitrary  constants. 

Ex.1.   Given  ^  =  I,findy. 
dx3      x3 

Integration  of  —  with  respect  to  x  gives 

dx1         2  & T     l' 
A  second  integration  gives, 

dx      2  x 
and  finally  y  =  \  log  x  +  \  Cix2  +  C2x  +  C8. 

EL.    CALC— 21  321 


322  INTEGRAL   CALCULUS 

The  triple  integration  required  in  this  example  will  be  symbolized  by 

which  will  be  called  the  triple  integral  of  —  with  respect  to  x. 

xs 

Ex.  2.  Determine  the  curves  having  the  property  that  the  radius 
of  curvature  at  any  point  P  is  proportional  to  the  cube  of  the  secant 
of  the  angle  which  the  tangent  at  P  makes  with  a  fixed  line. 

If  a  system  of  rectangular  axes  is  chosen  with  the  given  line  for 
x-axis,  it  follows  from  equation  (6),  p.  173,  and  from  Art.  42,  that 


\dxl  J       i  r1   ,(dyyii 
~y =«L    +UJ  J  ' 


dx2 
in  which  a  is  an  arbitrary  constant.     This  equation  reduces  to 

£*  =  «, 

dx* 

from  which  follows 

y  =  jjf[a(dxy  =  «[f  +  Cix  +  e2], 

Ci  and  Cz  being  constants  of  integration.     Hence  the  required  curves 
are  the  parabolas  having  axes  parallel  to  the  ?/-axis. 

The  existence  of  the  two  arbitrary  constants  Ci,  C2  in  the  preceding- 
equation  makes  it  possible  to  impose  further  conditions.  Suppose, 
for  example,  it  be  required  to  determine  the  curve  having  the  prop- 
erty already  specified,  and  having  besides  a  maximum  (or  a  minimum) 
point  at  (1,  0). 

Since  at  such  a  point  (-2L  =  0,  it  follows  that 
dx 

0  =  «(1+  Ci), 

whence  C\  =  —  1. 


SUCCESSIVE   INTEGRATION  323 

Also,  by  substituting  (1,  0)  in  the  equation  of  the  curve, 
0  =  a{\  -  1  +  CV), 


Accordingly  the  required  curve  is 

y=l(x-iy. 

Ex.  3.  Find  the  equation  (in  rectangular  coordinates)  of  the 
curves  having  the  property  that  the  radius  of  curvature  is  equal  to 
the  cube  of  the  tangent  length. 

[Hint.     Take  y  as  the  independent  variable.] 

Ex.  4.  A  particle  moves  along  a  path  in  a  plane  such  that  the 
slope  of  the  line  tangent  at  the  moving  point  changes  at  a  rate  pro- 
portional to  the  reciprocal  of  the  abscissa  of  that  point.  Find  the 
equation  of  the  curve. 

Ex.  5.  A  particle  starting  at  rest  from  a  point  P  moves  under  the 
action  of  a  force  such  that  the  acceleratiop  (cf.  Ex.  14,  p.  77)  at  each 
instant  of  time  is  proportional  to  (is  k  times)  the  square  root  of  the 
time.     How  far  will  the  particle  move  in  the  time  /? 

Ex.  6.  In  connection  with  a  certain  curve  referred  to  rectangular 
axes,  we  know  in  advance  that  it  passes  through  a  point  A  on  the 
y-axis  at  a  distance  1.12  in.  above  the  origin.  It  also  passes  through 
a  point  B  of  the  first  quadrant  which  is  at  a  distance  of  12  in.  from 
the  y-axis,  and  the  slope  of  the  tangent  to  the  curve  at  this  point  is 
0.09.  At  each  point  P  of  the  curve  the  second  derivative  of  y  satis- 
fies the  relation 

— {  =  0.0012  x. 

It  is  required  to  find  the  general  expression  (in  terms  of  x)  of  the 
ordinate  and  the  slope  of  the  tangent  line  for  any  point  P  of  the 
curve.      In  particular,  find  the  ordinate  and  slope  when  x  =  20  in. 


324  INTEGRAL   CALCULUS 

Ex.  7.  For  a  certain  curve  ADN  situated  in  the  first  quadrant  we 
have  given 

1000^  =  1.5  -0.276  x. 

dx2 

The  point  A  has  the  coordinates  (0,  0.04)  and  the  abscissa  of  D  is 
10.  At  the  point  B  of  the  curve,  whose  abscissa  is  5,  the  slope  of  the 
tangent  line  is  0.002. 

A  second  curve  DC  is  tangent  to  the  first  at  the  point  D,  and  for 
each  point  of  it  we  know  that 

1000  ^  =  0.2  x  -0.115. 
dx2 

Find  the  equations  of  both  curves. 

157.  Integration  of  functions  of  several  variables.  When 
functions  of  two  or  more  variables  are  under  consideration, 
the  process  of  differentiation  can  in  general  be  performed 
with  respect  to  any  one  of  the  variables,  while  the  others 
are  treated  as  constant  during  the  differentiation.  A  repeti- 
tion of  this  process  gives  rise  to  the  notion  of  successive 
partial  differentiation  with  respect  to  one  or  several  of  the 
variables  involved  in  the  given  function.     [Cf.  Arts.  62,  67.] 

The  reverse  process  readily  suggests  itself,  and  presents 
the  problem :  Given  a  partial  (first,  or  higher)  derivative  of  a 
function  of  several  variables  with  respect  to  one  or  more  of  these 
variables,  to  find  the  original  function. 

This  problem  is  solved  by  means  of  the  ordinary  processes 
of  integration,  but  the  added  constant  of  integration  has  a 
new  meaning.     This  can  be  made  clear  by  an  example. 

Suppose  u  is  an  unknown  function  of  x  and  y  such  that 

dx  J 


SUCCESSIVE   INTEGRATION  325 

Integrate  this  with  respect   to  x  alone,  treating  y  at  the 
same  time  as  though  it  were  constant.     This  gives 

u  =  x2  -f-  2  xy  -f-  <f>, 

in  which  <f>  is  an  added  constant  of  integration.  But  since 
y  is  regarded  as  constant  during  this  integration,  there  is 
nothing  to  prevent  <£  from  depending  on  it.  This  depend- 
ence may  be  indicated  by  writing  <f>(y)  in  the  place  of  <f>. 
Hence  the  most  general  function  having  2x  +  2y  for  its 
partial  derivative  with  respect  to  x  is 

u  =  x2  +  2xy+<f>(y), 

in  which  <f>(y)  is  an  entirely  arbitrary  function  of  y. 

Again,  suppose 

d2u  2  2 

=  xzy . 


dxdy 

Integrating  first  with  respect  to  y,  x  being  treated  as  though 
it  were  constant  during  this  integration,  we  find 

where  i{/(x)  is  an  arbitrary  function  of  x,  and  is  to  be  regarded 
as  an  added  constant  for  the  integration  with  respect  to  y. 

Integrate  the  result  with  respect  to  x,  treating  y  as  constant- 
Then 

u  =  i  «V  +  *0)  +  $>(?/). 

Here  ®(y),  the  constant  of   integration  with   respect  to  x, 
is  an  arbitrary  function  of  y,  while 

*(x)  =  J  ij/(x)dx. 

Since  if/(x)  is  an  arbitrary  function  of  x,  so  also  is  <l'(x). 


326  INTEGRAL   CALCULUS 

158.  Integration  of  a  total  differential.  The  total  differential 
of  a  function  u  depending  on  two  variables  has  been  defined 
(Art.  63)  by  the  formula 

,         du  7     .  du  - 

du  =  —  dx-\ ay. 

dx  dy 

The  question  now  presents  itself:  Given  a  differential  ex- 
pression of  the  form 

Pdx+Qdy,  (1) 

wherein  P  and  Q  are  functions  of  x  and  y,  does  there  exist 
a  function  u  of  the  same  variables  having  (1)  for  its  total 
differential  f 

It  is  easy  to  see  that  in  general  such  a  function  does  riot 
exist.  For,  in  order  that  (1)  may  be  a  total  differential  of  a 
function  u,  it  is  evidently  necessary  that  P  and  Q  have  the 
forms 

P=*     Q  =  p.  (2) 

dx  ay 

What  relation,  then,  must  exist  between  P  and  Q  in  order 
that  the  conditions  (2)  may  be  satisfied?  This  is  easily 
found  as  follows.  Differentiate  the  first  equation  of  2  with 
respect  to  y,  and  the  second  with  respect  to  x.     This  gives 

dP  =   d2u       dQ  =  d2u 
dy       dy  dx'     dx      dxdy 
from  which  follows  (Art.  68) 

8P=8Q  (3 

By       dx 

This  is  the  relation  sought. 

The  next  step  is  to  find  the  function  u  by  integration.  It 
is  easier  to  make  this  process  clear  by  an  illustration. 


SUCCESSIVE   INTEGRATION  327 

Given         (2  x  +  2y  +  2)dx  +(2y  +  2  x  +  2)dy, 

find  the  function  w  having  this  as  its  total  differential. 

Since         P=2x  +  2y  +  2,    Q=2y  +  2x  +  2, 
it  is  found  by  differentiation  that 

^=2  and   98-2, 

dy  ox 

hence  the  necessary  relation  (3)  is  satisfied. 
From  (2)  it  follows  that 

p  =  2x  +  2y  +  2. 
ox 

Integrating  this  with  respect  to  x  alone  gives 

u  =  x2  +  2xy  +  2x  +  <£(?/).  (4) 

It  now  remains  to  determine  the  function  <j>  (y)  so  that 

df[=Q]=2y  +  2x  +  2.  (5) 

Differentiating  (4)  with  respect  to  y  alone  gives 

1^  =  2*+^), 
dy 

where  <t>'(y)  denotes  the  derivative  of  <f>(y)  with  respect  to  y. 
The  comparison  of  this  result  with  (5)  gives 

2y  +  2x+2  =  2x  +  <l>\y)1 

or  <f>'(y)  =  2y  +  2,  (6) 

whence,  by  integrating  with  respect  to  y, 

4>(y)=y2  +  2y  +  C, 

in  which  O  is  an  arbitrary  constant    with  respect  to  both  x 
and  y. 

Hence  u  =  x2  +  2  xy  +  2  x  +  if  +  2  y  -f  C. 


328  INTEGRAL   CALCULUS 


EXERCISES 


Determine  in  each  of  the  following  cases  the  function  u  having  the 
given  expression  for  its  total  differential : 

1.  y  dx  +  x  dy. 

2.  sin  x  cos  y  dx  +  cos  x  sin  y  dy. 

3.  y  dx  —  x  dy. 

4     ydx-xdy 
xy 

5.    (3  x2  -  3  ay)dx  +  (3  y2  -  3  ax)  dy. 

g        y  dx  x  dy 

x'1  +  y'1      y'2  +  x'2 

7.  (2  x2  +  2xy  +  5)  dx  +  (x2  -\-  y2  -  y)  dy. 

8.  O4  +  i/4  +  x2  -  y2)  dx  +  (1  ysx  -  2  xy  +  y  -  y2  +  2)  dy.  <~ 

159.   Multiple  integrals.     The  integration  of  — —  was  con- 

dx  ay 

sidered  in  Art.  157.     If  F(x,  y)  is  written  for  the  given  func- 
tion, the  required  integration  will  be  represented  by  the  symbol 


=  ||  F(x,  y).dx  dy, 


and  the  function  sought  will  be  called  the  double  integral  of 
F(x,  y)  with  respect  to  x  and  y. 


Likewise  I  F{x,  y,  z)  dx  dy  dz 


will  be  called  the  triple  integral  of  F(x,  y,  z).     It  represents 

d3u 

the  function  u  whose  third  partial  derivative  —  is  the 

dx  dy  oz 

given  function  F(x,  y,  z).     It  will  be  understood  in  what  fol- 
lows that  the  order  of  integration  is  from  left  to  right,  that  is, 


SUCCESSIVE   INTEGRATION  329 

we  integrate  first  with  respect  to  the  left-hand  variable  x,  then 
with  respect  to  y,  and  lastly  with  respect  to  z. 

Such  integrals  (double,  triple,  etc.)  will  be  referred  to  in 
general  as  multiple  integrals. 

160.  Definite  multiple  integrals.  The  idea  of  a  multiple 
integral  may  be  further  extended  so  as  to  include  the  notion 
of  a  definite  multiple  integral  in  which  limits  of  integration 
may  be  assigned  to  each  variable. 

Thus  the  integral    I      I    x2ys  dy  dx  will  mean  that  x2)/  is  to 

be  integrated  first  with  respect  to  y  between  the  limits  0  and  2. 
This  gives 

►2 

x-y*  dy  =4  x2. 


j: 


The  result  so  obtained  is  to  be  integrated,  with  respect  to  x 
between  the  limits  a  and  b,  which  leads  to 


x 


h4:X2dx  =  ±(b3-a3) 


as  the  value  of  the  given  definite  double  integral. 
In  general  the  expression 


x'j; 


Fix,  y)dydx 


will  be  used  as  the  symbol  of  a  definite  double  integral.  It 
will  be  understood  that  the  integral  signs  with  their  attached 
limits  are  always  to  be  read  from  right  to  left,  so  that  in  the 
above  integral  the  limits  for  y  are  b  and  b',  while  those  for  x 
are  a  and  a'. 

Since  x  is  treated  as  constant  in  the  integration  with  re- 
spect to  y,  the  limits  for  y  may  be  functions  of  x.     Consider, 


330  INTEGRAL   CALCULUS 


for  example,  the  integral    J     J      xydydx.     The  first  integra- 
tion  (with  respect  to  y)  gives 


Jxy  dy  =  x 


,  x2      x\      Xs  —  X2 


By  integrating  this  result  with  respect  to  x  between  limits  0 
and  1  the  given  integral  is  found  to  have  the  value  —  ^. 

EXERCISES 

Evaluate  the  following  definite  integrals  : 

1.  f  *  ( 2xcos  (xy)dydx.  5-    j/jo  '    p2  sin  0  dp  dO. 

2.  jo  jo  x2  ^  <Zx.  6.    Jo  jy      Vx*,  -  y^  dx  dy. 

3      r«  fl°evdxdy  7      f 2  f «  Cxs/z  xdy  dxdz 

'    Ji  Jo  y     '  Ji  J<>  Jo  a;2  +  #2   * 


4 


'dzdydx 


.    J  j^sec2^),/,^.  8.    J(Jx      Jo       -+y  + 


161.  Plane  areas  by  double  integration.  The  area  bounded 
by  a  plane  curve  (or  by  several  curves)  can  be  readily  ex- 
pressed in  the  form  of  a  definite  double  integral.  An  illus- 
trative example  will  explain  the  method. 

Ex.  1.   Find  by  double  integration  the  area  of  the  circle 

(x  _  fl)2  +  (y  -  b)2  =  r2. 

Imagine  the  given  area  divided  into  rectangles  by  a  series  of  lines 
parallel  to  the  y-axis  at  equal  distances  Ax,  and  a  series  of  lines 
parallel  to  the  x-axis  at  equal  distances  A?/. 

The  area  of  one  of  these  rectangles  is  Ay  •  Ax.  This  is  called  the 
element  of  area.     The  sum  of  all  the  rectangles  interior  to  the  circle 


SUCCESSIVE   INTEGRATION 


331 


will  be  less  than  the  area  required  by  the  amount  contained  in  the 
small  subdivisions  which  border  the  circumference  of  the  circle. 

All  these  neglected  portions  are  contained  within  a  ring  bounded 
by  the  given  circle  and  a  circle  concentric  with  it,  whose  radius 
is  less  than  r  by  the 
length  of  diagonal  of 
an  element  of  area, 
that  is,  of  radius 


r  -  V(A;r)2  +  (Ay)K 
In  other  words,  the 
amount  neglected  is 
less  than  the  area  of  a 
circular  ring  whose 
width  is 


0 


V(Ax)2  +  (A*/)2 
and    which    therefore 

approaches  zero  simul-        !  r"    5 

taneously  with  Ax  and  Ay.     Hence  the  area  of  the  circle  is  the  limit 
of  the  sum  of  all  the  elements  of  area  included  within  it. 

To  find  the  value  of  the  limit  of  this  sum  it  is  convenient 
first  to  add  together  all  the  elements  contained  between  two  con- 
secutive parallels.  Let  PXP2  be  one  of  these  parallels  having  the 
direction  of  the  ar-axis.  Then  y  remains  constant  while  x  varies 
from  a  —  vV2  -  (y  —  b)2  (the  value  of  the  abscissa  at  P\)  to 
a  +  vV2  —(y  —  b)'1  (the  value  at  P2).  The  limit,  as  Ax  approaches 
zero,  of  the  sum  of  rectangles  in  the  strip  from  PiP2  is  evidently 


A#[limit  of  sum  (Ax  +  Ax  +••■)]  =  Ay  I 


a+vV-Cy-fi)2 
;-\/r2-(y-6)2 


dx.       (1) 


Now  find  the  limit  of  the  sum  of  all  such  strips  contained  within 
the  circle.  This  requires  the  determination  of  the  limit  of  the  sum 
of  terms  such  as  (1)  for  the  different  values  of  y  corresponding  to 
the  different  strips.     Since  y  begins  at  the  lowest  point  A  with  the 


332  INTEGRAL   CALCULUS 

value  b  —  r,  and  increases  to  b  +  r,  the  value  reached  at  B,  the  final 
expression  for  the  area  is 

rb+r  i-  /*a+v'r2-(2/-&)2  ~1  fb+r    ra+\Zr^-(y~b^ 

\  \       . .  dx  \dy  =  \        \ ,  dx dy. 

Jb-r     LJa-VrZ-(y-b)*  J  Jh-r    ^a-Vr2-(y-b)2 

Integrating  first  with  respect  to  x  gives 

e-w*=s=ir        i>^p^=  2Vr,_(y_6)2, 

Ja-vV2-(y-&)2  Ja-\/»*-(y-6)'  w  ' 

This  result  is  then  integrated  with  respect  to  y,  giving 


f>+r2Vr2-  (y  -  b)2dy  =  (y-b)Vr2-  (y  -  by  +  r2 


sm  ^ —irr 


r 


J»- 


If  the  summation  had  begun  by  adding  the  rectangles  in  a  strip 
parallel  to  the  y-axis,  and  then  adding  all  of  these  strips,  the  expres- 
sion for  the  area  would  take  the  form 


)         )       , dydx. 


It  is  seen  from  the  last  result  that  the  order  of  integration  in  a 
double  integral  can  be  changed  if  the  limits  of  integration  are  properly 
modified  at  the  same  time. 

Ex.  2.    Find  the  area  which  is  included  between  the  two  parabolas 

y2  =  9  x  and  y2  =  72  -  9  x. 

Ex.  3.    Find  the  area  between  y2  =  5a;  and  y  =  x. 

Ex.  4.  Find  by  double  integration  the  area  of  the  segment  of  the 
circle  x2  +  y2  =  16  cut  off  by  the  line  x  +  y  =  4. 

Ex.  5.    Find  the  area  between  the  two  curves 
y8  =  x  and  y  =  xz. 

Ex.  6.   Find  the  area  between  the  two  curves 
y2  =  xs  and  y2  =  x. 

Ex.  7.    Find  by  double  integration  the  area 
_J2     of  one  loop  of  the  polar  curve  p  =  a  sin  2  0. 
Fi<;.  W  Imagine  the  area  divided  into   small  ele- 


SUCCESSIVE   INTEGRATION  333 

merits  by  means  of  concentric  circles  whose  radii  vary  by  equal 
increments  Ap  and  by  means  of  radii  drawn  from  the  origin,  the 
angle  between  two  consecutive  radii  being  A0.     (See  Fig.  86.) 

The  area  of  an  arbitrary  element  may  be  expressed  as  the  differ- 
ence of  two  circular  sectors  with  a  common  angle  A0  and  with  radii 
p  +  Ap  and  p  respectively.     That  is, 
element  of  area  =  \(p  +  Ap)2  A0  -  \  p2  A0 
=  pA0Ap  +  iA0(Ap)2. 

The  sum  of  all  the  complete  elements  within  the  loop  may  then  be 
represented  by  the  formula 

^pA0Ap  +  i^A0(Ap)2. 

Reasoning  precisely  as  in  Ex.  1,  we  find  the  limit  of  the  first  sum 
to  be 

r\  /-a  sin  2  0 

The  second  sum  may  be  written  \  Ap2  A0  Ap,  hence  its  limit  is 

n 

I  •  lira  Ap  •  lim  ^A(9  Ap=  i  *  °  *  J0 ~  Jo  dP  dB  =  °' 

Following  the  analogy  of  Ex.  1,  we  can  easily  see  that  all  the 
neglected  incomplete  elements  of  area  lie  within  a  narrow  band  along 
the  boundary  of  the  given  area,  the  width  of  which  band  approaches 
0.  Their  sum  therefore  approaches  zero  in  passing  to  the  limit. 
•  It  follows  from  the  preceding  discussion  that  the  general  formula 
for  area  in  polar  coordinates  is 


§§pdpdO, 


the  limits  of  integration  being  determined  by  the  boundary  of  the 
given  area. 

Ex.  8.    Find  by  double  integration  the  area  of  the  cardioid 

p  =  a(l  —  cos  0). 


334  INTEGRAL   CALCULUS 

Ex.  9.   Find  the  area  of  the  lemniscate  p2  =  a2  cos  2  0. 

Ex.  10.  Express  by  double  integrals  the  three  areas  between  the 
cardioid  (Ex.  8)  and  the  circle  p  =  a. 

Ex.  11.  Find  by  double  integration  the  area  of  the  triangle  whose 
vertices  have  the  rectangular  coordinates  (5,  2),  (—3,  6),  (7,  6). 

Ex.  12.   Find  the  area  common  to  the  two  circles 

x1  -  8  x  +  y2  -  8  y  +  28  =  0, 
x2-  8a:+  y1  -  ±y  +  16  =.0. 

162.  Volumes.  The  volume  bounded  by  one  or  more  surfaces 
can  be  expressed  as  a  triple  integral  when  the  equations  of  the 
bounding  surfaces  are  given. 

Let  it  be  required  to  find  the  volume  bounded  by  the  surface 
ABC  (Fig.  87)  whose  equation  is  z—f(x,  y),  and  by  the  three 
coordinate  planes. 

Imagine  the  figure  divided  into  small  equal  rectangular 
parallelopipeds  by  means  of  three  series  of  planes,  the  first 
series  parallel  to  the  ?/z-plane  at  equal  distances  Ax,  the  second 
parallel  to  the  ccz-plane  at  equal  distances  Ay,  and  the  third 
parallel  to  the  a?2/-plane  at  equal  distances  Az.  The  volume 
of  such  a  rectangular  solid  is  Ax  Ay  Az ;  it  is  called  the  element 
of  volume.  The  limit  of  the  sum  of  all  such  elements  con- 
tained in  OABC  is  the  volume  required,  provided  that  the 
bounding  surface  ABC  is  continuous.  For  the  sum  of  the 
neglected  incomplete  elements,  which  border  the  surface,  is 
less  than  the  volume  of  a  shell  whose  outside  boundary  is 


the  given  surface  and  whose  thickness  is  V(A#)2-f-(Ay)2-f-(Az)2, 
the  diagonal  of  the  element  of  volume.  Hence  the  error  ap- 
proaches zero  as  the  three  increments  diminish. 

To   effect  this  summation,  add  first  all  the   elements  in  a 


SUCCESSIVE   INTEGRATION 


33; 


vertical  column.  This  corresponds  to  integrating  with  respect 
to  z  (x  and  y  remaining  constant)  from  zero  to  f(x,  y).  ,  Then 
add  all  such  vertical  columns  contained  between  two  consecu- 
tive planes  parallel  to  the  ?/z-plaiie  (x  remaining  constant), 
which  corresponds  to  an  integration  with  respect  to  y  from 
y  =  0  to  the  value  attained  on  the  boundary  of  the  curve  AB. 


Fig  87. 


This  value  of  y  is  found  by  solving  the  equation  f(x,  y)  =  0. 
Finally,  add  all  such  plates  for  values  of  x  varying  from  zero 
to  its  value  at  A.     The  result  is  expressed  by  the  integral 


XX  X    dzd'Jdx> 


336  INTEGRAL   CALCULUS 

in  which  <f>(x)  is  the  result  of  solving  the  equation  f(x,  y)  =  0 
for  y,  and  a  is  the  ^-coordinate  of  A. 

Ex.  1.    Find  the  volume  of  the  sphere  of  radius  a. 
The  equation  of  the  sphere  is 

x2  +  ?/2  +  z%  _  a2? 


or  z  =  Va2  —  x2  —  y'K 

Since  the  coordinate  planes  divide  the  volume  into  eight  equal 
portions,  it  is  sufficient  to  find  the  volume  in  the  first  octant  and 
multiply  the  result  by  8. 

The  volume  being  divided  into  equal  rectangular  solids  as  described 
above,  the  integration  with  respect  to  z  is  equivalent  to  finding  the 
limit  of  the  sum  of  all  the  elements  contained  in  any  vertical  column. 
The  limits  of  the  integration  with  respect  to  z  are  the  values  of  z 
corresponding  to  the  bottom  and  the  top  of  such  a  column,  namely, 
z  —  0,  and  z  =  Va2  —  x2  —  y2,  since  the  point  at  the  top  is  on  the  sur- 
face of  the  sphere. 

The  limits  of  integration  with  respect  to  y  are  found  to  be  y  =  0 
(the  value  at  the  x-axis),  and  y  —  Va2  —  x2  (the  value  of  y  at  the  cir- 
cumference of  the  circle  a2  —  x2  —  y2  =  0,  in  which  the  sphere  is  cut 
by  the  a^-plane). 

Finally,  the  limiting  values  for  x  are  zero  and  a,  the  latter  being 
the  distance  from  the  origin  to  the  point  in  which  the  sphere  inter- 
sects the  x-axis.     Hence 

V[=  volume  of  sphere]  =  8  j^ £ V"2-*2 ^V"2 -**-** dz  dy  dx. 

Integration  with  respect  to  z  gives 

V  =  8  i     \  Va2  -  x2  -  y2  dy  dx; 

then  with  respect  to  y  and  x, 

V  =  8  j>  [|  V^^V,  +  ^^-^-J^ 


SUCCESSIVE   INTEGRATION  337 

Ex.  2.   Find  the  volume  of  one  of  the  wedges  cut  from  the  cylinder 
x2  +  y2  =  a2  by  the  planes  2  =  0  and  z  =  mx. 

Ex.  3.    Find  the  volume  common  to  two  right  circular  cylinders 
of  the  same  radius  a  whose  axes  intersect  at  right  angles. 

Ex.  4.    Find  the   volume  of   the    cylinder    (x  —  1)2+  (y  —  l)2  =  1 
limited  by  the  plane  z  =  0,  and  the  hyperbolic  paraboloid  z  =  xy. 

Ex.  5.   Find  the  volume  of  the  ellipsoid 
a2      b2      c2 


Ex.  6.    Find  the  volume  of  that  portion  of  the  elliptic  paraboloid 

z=l-*-£ 
a2      b2 

which  is  cut  off  by  the  plane  z  =  0. 

Ex.  7.    Find  by  triple  integration  the  volume  of  the  tetrahedron 
formed  by  the  three  coordinate  planes  and  the  plane  x  +  2  y  +  3z  —  1. 

Ex.  8.    Find  the  volume  of  the  elliptic  paraboloid  2  y2  -f  3  z'1  —  6  x 
cut  off  by  the  plane  x  =  2. 


el.  calc.  — 22 


Jf- 


CHAPTER   IX 

SOME  APPLICATIONS  OF  INTEGRAL  CALCULUS  TO 
PROBLEMS  OF  MECHANICS 

163.    Liquid  pressure  on  a  plane  vertical  wall.     The  pressure 
exerted  by  the  liquid  upon  any  point  of  a  plane  vertical  wall 

is   proportional    to   the    depth 

Surface B 

7       of   that   point    below    the    sur- 

f  /        face  of  the  fluid.     To  calculate 

\    i       y 

the   pressure    upon    the    entire 

wall    we    divide    it    into    nar- 
row horizontal  strips  of  equal 
areas  Ai.    Denote  the  breadth 
FlG-  88'  of  the  Jcth  strip  PQ  (Fig.  88), 

counting  from  the  top,  by  hk.  The  pressure  exerted  on  the 
ftth  strip  is  equivalent  to  the  weight  of  a  column  of  fluid 
standing  on  a  base  of  the  same  area  AA  and  having  an 
altitude  intermediate  between  the  least  depth  x  and  the 
greatest  depth  x  +  hk  of  points  on  the  given  strip.  This 
altitude  may  be  represented  by  x  4-  0khk  in  which  0k  has  a 
value  between  0  and  1.  If  w  denotes  the  weight  of  a  cubic 
unit  of  the  fluid,  the  pressure  on  PQ  is  w(x  +  0khk)  Ai. 
Summing  the  pressures  for  all  the  strips  of  the  wall,  we 
obtain  for  the  total  pressure 

^w(x  +  6khk)  AA. 
338 


SOME   APPLICATIONS   TO    MECHANICS  339 

In  order  to  evaluate  this  sum  we  take  its  limit  as  &A 
approaches  zero.  This  gives,  by  separating  into  two  terms 
and  observing  that  w  is  constant, 


Km  yx&A  +  w  lim  V  #A  A  A 


w 

A.4=b0 


The  second  term  reduces  to  zero.     For, 

]£0A  A.4  =  AA^ekhk<  ^A  •  H  (since  6k  <  1), 

in  which  H  denotes  the  total  altitude  of  the  wall ;  as  A^l  =  0 
the  right  member  of  this  inequality  approaches  zero.     Hence 


pressure  =  iv  I  x  dA. 


In  order  to  evaluate  the  integral,  it  is  most  convenient  to 

make  x  the  variable  of  integration.     Denote  by  y  the  width  of 

the  wall  at  the  depth  x.     Then  &A  =  yk  Ax  in  which  yk  is  a 

certain  value  of  y  between  y  and  2/+A?/.     (Compare  Art.  40.) 

Dividing  by  A£  and   passing   to   the   limit   we  obtain,  since 

lim  yk  =  y, 

dA=     dx 

dt  ~V  dt ' 

or  in  the  differential  notation,  dA  —  y  dx.     The  substitution  of 
this  in  the  above  integral  gives 


pressure 


=  w  j  xy  dx, 


the  limits  of  integration  being  the  values  of  x  at  the  top  and 
the  bottom  of  the  given  wall  or  surface. 

If  the  liquid  is  water  and  the  unit  of  length  is  a  foot,  then 


w  =  62^  lb. 


340  INTEGRAL   CALCULUS 

EXERCISES 

1.  Find  the  pressure  on  the  end  of  a  rectangular  tank  full  of  water 
that  is  10  ft.  long,  8  ft.  wide,  and  5  ft.  deep. 

2.  A  watermain  6  ft.  in  diameter  is  half  full  of  water.  Find  the 
pressure  on  the  gate  that  closes  the  main. 

3.  A  vertical  masonry  dam  in  the  form  of  a  trapezoid  is  200  ft. 
long  at  the  surface  of  the  water,  150  ft.  long  at  the  bottom,  and  60 
ft.  high.     What  pressure  must  it  withstand  ? 

4.  A  vertical  cross  section  of  a  trough  is  a  parabola  with  vertex 
downwards,  the  latus  rectum  lying  in  the  surface  and  being  4  ft. 
long.  Find  the  pressure  on  the  end  of  the  trough  when  it  is  full  of 
water. 

5.  One  end  of  an  unfinished  watermain,  4  ft.  in  diameter,  is  closed 
by  a  temporary  bulkhead  and  the  water  is  let  in  from  the  reservoir. 
Find  the  pressure  on  the  bulkhead  if  its  center  is  40  ft.  below  the 
surface  of  the  water  in  the  reservoir. 

164.    Center  of  gravity.     (1)  For  a  system  of  n  particles.     Let 
P1?  P2  be  two  particles  of  matter  of  masses  (or  weights)  mx  and 
p     m2,  respectively,  and  let  xlf  x2  be  their 
O        "       f  *~~   distances  from  a  chosen  point  O  on  the 

Fig.  89  straight     line    through    them.      There 

exists  a  point  P  such  that  the  segments  PXP  and  PP2  are  in- 
versely proportional  to  the  masses  of  the  two  points,  that  is, 

PiP=  %_  ^ 

PP2     %' 

Let  x  represent  the  distance  OP.    Then  formula  (1),  expressed 
in  terms  of  the  abscissas  of  the  points,  is 
x  —  xl_m2 


SOME   APPLICATIONS   TO   MECHANICS  341 

whence,  by  solving  for  x, 

mlx1-\-m2x2  /ON 

m!  +  m2 

The  point  P  is  called  the  center  of  gravity,  or,  the  center  of 
mass,  of  the  system  formed  by  the  two  points  Ply  P2.  If  we 
imagine  the  line  PXP2  to  consist  of  a  rigid,  weightless  rod 
with  the  two  given  particles  fastened  at  its  extremities,  and  if 
we  suppose  this  object  to  rest  on  the  point  P  as  a  base,  it  will 
remain  in  equilibrium,  without  any  tendency  in  either  of  the 
end  points  to  move  downward  under  the  force  of  gravity. 

In  other  words,  the  system  of  two  particles  is  equivalent,  as 
far  as  the  action  of  gravity  is  concerned,  to  a  single  particle, 
of  mass  mx  -+-  m2,  placed  at  the  point  P. 

Let  P3  be  a  third  point  of  mass  m3  situated  on  the  same  line 
with  Px  and  P2.  Then  the  abscissa  x  of  the  center  of  gravity 
of  the  system  of  three  points  may  be  found  by  calculating  the 
center  of  gravity  of  the  pair  P3  and  P  (the  center  of  gravity 
for  Plf  P2),  the  mass  of  P  being  taken  as  m1  -f  ra2,  the  sum  of 
the  masses  of  Px  and  P2.     This  gives 

Ox  +  m2)    1  l  ; — =2  +  m3x3 

w&i  +  m2  m^  4-  m2x2  -f-  m3X3m 

(mx  -+-  WI2)  +  ms  wh  +  m2  +  nh 


x  = 


In  like  manner  the  center  of  gravity  for  any  number  n  of 
particles  situated  on  a  straight  line  is  given  by  the  formula 

-  _  m^  +  m2x2  -{ +  mnxn^  ,^ 

m1  +  m2  +  •••  +mn 

If  the  n  particles  are  not  on  a  straight  line  but  are  situated 
in  the  same  plane  at  the  points  (a^,  y^),  (x2,  y2),  •  ••,  (xn1  yn), 


342  INTEGRAL    CALCULUS 

then  the  center  of  gravity  of  the  system  has  its  abscissa  given 
by  (3)  and  its  ordinate  y  is 


y  =  WW  +  m2y2  + h  m^ri 

m^mz-l \-mn 


If  the  n  particles  are  not  situated  in  one  plane,  there  will 
be  a  third  and  similar  formula  for  z. 

(2)  For  a  continuous  solid.  Imagine  the  solid  divided  up 
into  small  elements,  precisely  as  in  determining  its  volume,  by 
means  of  three  series  of  planes  parallel  to  the  coordinate 
planes  and  at  distances  Ax,  Ay,  Az.  If  we  regard  any  par- 
ticular element  as  being  very  nearly  of  uniform  density,  then 
the  mass  of  an  arbitrary  element  is  approximately  p  Ax  Ay  Az, 
in  which  p  is  the  weight  of  a  cubic  unit  of  homogeneous  mat- 
ter having  the  same  density  as  the  given  element.  This  num- 
ber p  is  usually  called  the  density.  For  a  finite  number  of 
elements  the  ^-coordinate  of  the  center  of  gravity  is  determined 
approximately  by  (3)  in  the  form 


(pigq  +  P1F2  +  —  +  P  A)  Aa;  Ay  Az 
0>i  +  p2+  >~  +  pn)AxAyAz 


in  which  xly  x2,  ••  are  the  abscissas  for  the  different  elements 
and  plt  p2,  •••  are  their  densities.  The  abscissa  of  the  center  of 
gravity  of  the  given  continuous  solid  is  obtained  by  making 
Ax,  Ay,  Az  approach  zero  as  a  limit.*     This  gives 


J   I    I  pxdxdydz 
I    I    Ipdxdydz 


*  A  proof  of  this  statement  will  be  found  in  Art.  166. 


SOME   APPLICATIONS  TO    MECHANICS  343 

the  limits  of  integration  being  determined  just  as  in  calculat- 
ing the  volume  of  the  solid.  If  the  solid  is  homogeneous, 
p  is  constant  and  cancels  out  of  numerator  and  denominator. 
Otherwise,  it  is  a  function  of  x,  y,  z. 

In  precisely  the  same  manner  the  values  of  y  and  z  are 
obtained.  The  coordinates  of  the  center  of  gravity  are  thus 
found  to  be 


x 

M 


m  j  j  j9  x  dx  dy  dz' y = mSSSp  y  dx  dy  dz' 
'^SSSpzdxdydZi 

in  which  p  is  the  density  at  the  point  (x,  y,  z)  and  M  is  the 
total  mass  of  the  given  solid,  that  is, 

M=  J   J   (pdxdydz. 

The  coordinates  of  the  center  of  gravity  of  a  plane  area  are 
found  in  like  manner  to  be 

x  =  —  j Jpxdxdy,  T/  =  ~JJpydxdy,  M=Jjpdxdy. 

EXERCISES 

In  the  following  problems  p  is  understood  to  be  constant  unless 
otherwise  specified.  The  abbreviation  C.  G.  will  be  used  for  "  center 
of  gravity." 

1.  Find  the  C.  G.  of  the  tetrahedron  whose  faces  are  the  three 
coordinate  planes  and  the  plane  x  +  2y  +  3z  =  6. 

2.  Find  the  C.  G.  of  the  volume  bounded  by  the  coordinate  planes 
the  plane  x  -f  y  =  1,  and  the  surface  z  =  xy. 


344  INTEGRAL   CALCULUS 

3.  Find  the  C.  G.  of  the  volume  bounded  by  the  hyperboloid 

- —  ^ —  —  =  1  and  the  plane  x  =  k,   k  >  a. 
a*     b2     c2  r 

4.  Find  the  C.  G.  of  the  semiellipsoid  on  the  positive  side  of  the 

xy-plane,  the  equation  of  the  ellipsoidal  surface  being  —  -f  ^-  -\ —  =  1. 

a2      b2      c2 

5.  Find  the  C.  G.  of  a  thin  hemispherical  shell  of  thickness  h 
bounded  by  two  concentric  hemispheres  of  radii  a  and  a  +  h. 

6.  A  hemispherical  iron  bowl  of  uniform  thickness  a  is  filled  with 
water.  If  the  density  of  iron  is  seven  times  that  of  water,  find  the 
C.  G.,  supposing  the  radius  of  the  interior  of  the  bowl  to  be  r. 

[Hint.  Find  the  C.  G.  of  the  iron  bowl  by  means  of  Ex.  5.  Find 
the  C.  G.  of  the  hemisphere  of  water  and  combine  the  centers  of  grav- 
ity of  the  iron  and  the  water  by  means  of  (2).] 

7.  Show  that  the  C.  G.  of  a  triangular  plate  one  inch  thick  is  one 
half  inch  below  the  intersection  of  the  medians  of  the  upper  face. 

8.  Find  the  C.  G.  of  a  T-iron  one  inch  thick,  the  vertical  bar  being 
a  inches  wide  and  b  inches  high,  and  the  horizontal  bar  a'  inches  wide 
and  V  inches  long. 

9.  Find  the  C.  G.  of  a  sector  of  a  circle  of  radius  a  and  angle  0. 

10.  Find  the  C.  G.  of  the  segment  of  the  circle  x2  +  y*  =  r2  cut  off 
by  the  line  x  =  a,  0  <  a  <  r. 

11.  Find  the  C.  G.  of  the  quadrant  of  an  ellipse. 

12.  Find  the  C.  G.  of  the  segment  of  an  ellipse  cut  off  by  the 
chord  joining  the  extremities  of  the  major  and  minor  axes. 

13.  Find  the  C.  G.  of  the  area  bounded  by  the  parabola 

Vx  +  y/y  =  Va 
and  the  line  x  +  y  =  a. 

14.  Prove  that  the  volume  of  a  solid  of  revolution  is  equal  to  the 
product  of  the  generating  area  by  the  length  of  path  described  by  its 
center  of  gravity. 


SOME   APPLICATIONS   TO   MECHANICS  345 

15.  Find  the  C.  G.  of  an  octant  of  an  ellipsoidal  mass. 

16.  Find  the  C.  G.  of  the  preceding  mass  when  the  density  varies 
directly  as  the  distance  from  the  plane  x  =  0. 

17.  Find  the  C.  G.  of  an  octant  of  a  sphere.  From  this  result  find 
the  C.  G  of  an  octant  of  a  spherical  shell  of  thickness  h  and  inner 
radius  a. 

18.  Find  the  C.  G.  of  an  octant  of  a  sphere  if  the  density  varies 
directly  as  the  distance  from  the  center  of  the  sphere. 

[Hint.  Divide  up  into  thin  concentric  shells  of  equal  thickness  h, 
the  density  of  a  particular  shell  being  regarded  as  constant.  Let  A. 
denote  the  radius  of  an  arbitrary  shell,  X  the  distance  of  its  C.  G.  from 
the  origin,  and  m  its  mass.  Calculate  X  in  terms  of  A  by  means  of 
Ex.  17,  measuring  it  on  a  line  equally  inclined  to  the  x,  y,  z  axes. 
Then  use  the  different  values  of  X  in  place  of  xi,  X2,  ••• ,  formula  (3), 
and  pass  to  the  limit. 

19.  Find  the  C.  G.  of  a  right  circular  cone  of  altitude  h  and  base- 
radius  r. 

This  problem  can  be  solved  by  single  integration  if  we  suppose  the 
solid  divided  up  into  thin  plates  of  equal  thickness  by  means  of 
planes  parallel  to  the  base.  Then  find  the  approximate  expression 
for  the  C.  G.  of  any  plate,  apply  (3),  and  pass  to  the  limit. 

20.  Find    the    C.  G.    of     the     portion     of     the    elliptical    cone 

—a  +  f-  =  (z  —  l)2  between  the  vertex  (0,  0,  1)  and  the  zv-plane. 
a2      b2 

21.  A  cone  of  vertical  angle  2  0  has  its  vertex  on  the  surface  of  a 
sphere,  its  axis  passing  through  the  center  of  the  sphere. 

(a)  Find  the  C.  G.  of  the  mass  outside  the  cone  and  inside  the 
sphere. 

(b)  Find  the  C.  G.  of  the  mass  inside  the  sphere  and  inside  the 
cone. 


346  INTEGRAL   CALCULUS 

165.  Moment  of  Inertia.  The  moment  of  inertia  of  a  small 
particle  of  matter  of  mass  m  about  an  axis  is  defined  as  the 
product  of  the  mass  by  the  square  of  the  distance  of  the 
particle  from  the  axis.  It  measures  the  resistance  of  the  par- 
ticle to  rotation  about  the  axis. 

To  find  the  moment  of  inertia  of  a  homogeneous  solid  body, 
imagine  it  divided  up  into  small  rectangular  blocks  (or  ele- 
ments) of  dimensions  Asc,  Ay,  Az.  Then  the  moment  of 
inertia  of  a  single  element  about  the  x'-axis  is  approximately 

p(y2  +  z~)&x  ty  Az, 
in  which  p  is  the  density,  that  is,  it  is  the  weight  of  a  cubic 
unit  of  the  given  solid.     Summing  up  these  elements  over  the 
whole  body  and   taking  the   limit   of  the   sum,  we  find   the 
moment  of  inertia  to  be  * 


///< 


p(y2  +  z2)dxdydzf  (4) 

the  triple  integral  being  extended  over  the  entire  solid,  just 
as  was  done  in  finding  its  volume. 

If  the  solid  is  not  homogeneous,  then  p  is  variable.  Its 
value  at  a  specified  point  P  of  the  given  body  is  equal  to  the 
weight  of  a  homogeneous  cubic  unit  of  matter  having  the 
same  density  throughout  as  the  particle  of  matter  at  the  point 
P.  It  is  a  function  of  x,  y,  z  which  is  to  be  determined  by  the 
conditions  of  the  given  problem. 

Similarly,  the  moment  of  inertia  of  a  plane  area  about  the 
a>axis  is  defined  as  the  limit  of  the  sum  of  terms  formed  by 
multiplying  each  element  of  area  by  the  square  of  its  distance 
from  the  axis.     This  gives  the  formula 

y2  dx  dy. 


//• 


*  See  the  next  article  for  a  completion  of  the  proof. 


SOME   APPLICATIONS   TO   MECHANICS  347 

EXERCISES 

In  the  following  problems  M.I.  is  used  for  brevity  to  denote 
'••  moment  of  inertia."  Unless  the  contrary  is  stated,  the  body  is 
homogeneous  and  of  density  p. 

1.  Find  the  M.  I.  of  a  rectangular  parallelopiped  of  dimensions  a, 
b,  c  about  an  edge  a. 

Take  three  edges  a,  b,  c  meeting  in  a  common  point  as  the  x,  y,  z 
axes,  respectively.     Then  by  formula  (4)  the  M.  I.  is 


"!oTJ>+-~2)*"-yrf*- 


2.  Find  the  M.  I.  of  a  circular  cylinder  of  radius  a  and  altitude  h 
about  its  axis. 

3.  Find  the  M.I.  of  the  cylinder  of  Ex.  2  about  a  line  perpendicu- 
lar to,  and  bisecting,  the  axis. 

4.  Find  the  M.  I.  of  a  circular  cone  of  altitude  a  and  base-radius 
r  about  its  axis. 

Hint.    If  the  axis  of  the  cone  is  taken  for  the  x-axis  and  its  vertex 
at  the  origin,  the  equation  of  the  conical  surface  is 

x2  _  y2  +  z2  "I 
a2  r2      J 

5.  Find  the  M.  I.  of  an  elliptical  right  cylinder  about  its  longi- 
tudinal axis,  the  axes  of  the  elliptical  bases  being  2  a,  2  b  and  the 
altitude  h. 

6.  Find  the  M.I.  of  the  preceding  solid  about  the  minor  axis  of 
an  elliptical  base. 

7.  Find  the  M.I.  of  the  same  body  about  a  line  bisecting  the 
longitudinal   axis    and    parallel   to  the  major  axes  of   the  elliptical 


348  INTEGRAL   CALCULUS 

8.  Find  the  M.  I.  of  a  sphere  about  a  diameter.  Hence  find  the 
M.  I.  of  a  spherical  shell  of  uniform  thickness  h  about  a  diameter,  as- 
suming that  the  M.  I.  of  a  solid  consisting  of  two  parts  is  the  sum  of 
the  moments  of  the  separate  parts. 

9.  Find  the  M.I.  of  a  spherical  solid  of  radius  r  about  a  diameter 
if  the  density  varies  directly  as  the  nth  power  of  the  distance  from 
the  center. 

[Hint.  Imagine  the  sphere  divided  into  concentric  shells  of  equal 
thickness  AA  and  denote  by  X  the  interior  radius  of  any  shell.  Using 
the  preceding  problem,  write  down  the  element  of  M.  L,  that  is,  the 
M.I.  of  the  shell  of  radius  X  and  thickness  AA..  Take  the  limit  of 
the  sum  of  all  such  elements  as  AA.  =  0.  The  required  M.  I.  is  thus 
obtained  by  a  single  integration.] 

10.  Find  the  M.  I.  of  a  cube  of  edge  a  about  its  diagonal. 
[Hint.     Take  three  faces  of  the  cube  as  coordinate  planes.     Obtain 

an  expression  for  the  square  of  the  distance  from  any  point  (#,  y,  z) 
to  the  diagonal  of  the  cube  that  passes  through  the  origin.  This, 
multiplied  by  AxAyAz,  will  be  the  element  of  M.I.  Then  take  the 
limit  of  the  sum.] 

11.  Find  the  M.  I.  of  a  cylindrical  shell,  of  length  a,  about  its 
axis,  the  radius  of  the  inner  surface  being  r  and  that  of  the  outer 
surface  being  R. 

12.  Find  the  M.  I.  of  a  rectangle  of  sides  a,  b  about  the  side  b. 

13.  Find  the  M.  I.  of  a  triangle  of  base  b  and  altitude  h  about  an 
axis  through  a  vertex  parallel  to  the  opposite  side. 

14.  Find  the  M.  I.  of  a  circle  of  radius  a  about  a  diameter. 

166.  Duhamel's  Theorem.  In  order  to  complete  the  proof 
of  the  formulas  for  center  of  gravity  and  moment  of  inertia, 
we  make  use  of  the  following  theorem  which  is  of  very  general 
use  in  applications  of  the  Integral  Calculus. 


SOME   APPLICATIONS   TO   MECHANICS  349 

Duhamel's  Theorem.  Let  aly  ol>,  •••,  an  be  positive  variables, 
each  of  which  approaches  zero  as  n  increases  without  limit,  and 
suppose  that  the  sum  ai  +  c^-j-  •••  +  «n  approaches  a  finite  limit 
as  n  =  oc .  Let  /3i,  (32,  •••,  ft  be  variables  having  the  same  prop- 
erty as  the  a's  and  such  that  ™  ^  =  1  for  k=l,  2,  •  •-,  n. 
Then  "* 

^(ft  +  ft*  -  +«  =  n1^oo(«1  +  «2+  ...+«B). 

Since      lim    &  =  1,  we  may  write   &   in  the  form  l  +  ek  in 
«  =  «>«*  «fc 

which  ek  approaches  zero  as  n  =  oc.     Hence, 
ft  =  ak  +  et%, 

and  therefore  2  ft  =  S  "*  +  S  **<**• 

Let  e  denote  the  positive  value  of   the   numerically  greatest 
term  of  the  series  ey,  c2,  •••,  e„.     Then  we  have  the  inequalities 

—  €a1'^€lal<  4-««u 

—  ca2 5^  e2a2  ^  4-  ca2 , 


and  by  adding  we  obtain 

—  «(«!  +  «o  4-  •  •  •  4-  «„) ^  2e*«*  ^  +  «(«i  4-  «2  4- h  <*»)• 

Now  let  n  increase  without  limit.  Since  by  hypothesis  e  =  0 
and  («!4-«2+  •••  4-  an)  has  a  finite  limit,  it  follows  that  the 
first  and  last  members  of  the  preceding  inequalities  vanish  at 


350  INTEGRAL   CALCULUS 

the  limit  and  therefore      . 

Hence  Jf^A- J?.S*# 

As  an  application  of  the  above  theorem,  consider  the  sum 
occurring  in  the  approximate  formula  for  center  of  gravity, 

nam6ly'  <«%  +  *%+•••+ PA)  AF 

in  which  AF=  A#  A?/  Az. 

Let  pk,  xk  be  the  minimum,  and  pk",  xk"  the  maximum  values 
of  p,  x  in  the  ftth  element  of  volume.     For  brevity  write 

Pk'xk'*V=ak,   Pk"xk''±V=(3k. 
Then  we  have  ak  <^  pkxkAV^  /3k 

hence,  by  taking  the  sum, 

6      o  "x  " 
But    £-*  =  ^*    *     which  approaches  1  as  n  increases  since  pk',  xk 

approach  equality  with  pk",  xk".     Hence 

lim  Xa*  =  lim  ]£  A  =  lira  ^pa  AF. 

In  obtaining  this  result  no  restriction  is  placed  on  xk  and  pk 

*  A  variable  which  has  zero  as  a  limit  is  often  called  an  infinitesi- 
mal.   Hence  ai,  a2,  •••,  an  are  infinitesimals.    If  we  write   Sj.  =  e^,   then 

„  V^L  —  =  lim  ek  =  0.    When  two  infinitesimals,  5  and  a,  are  so  related  that 

the  ratio  of  5  to  a  has  the  limit  zero,  then  5  is  said  to  be  infinitesimal  with 
respect  to  a,  or  it  is  called  an  infinitesimal  of  a  higher  order  than  a. 

Since,  by  Duhamel's  Theorem,  lim  ^(a*.  -f  5*)  =  lim  2a*i  tn*8  theorem  is 
equivalent  to  saying  that  the  limit  of  a  sum  of  infinitesimals  is  not  affected 
by  dropping  from  each  term  an  infinitesimal  of  a  higher  order. 


ANSWERS 


DIFFERENTIAL   CALCULUS 

Page  23.     Art.  7 
2x-2;  2;  0;  1.  6.   wx"-1. 


2.   6x-4. 


3. 

1 
4  a* 

4. 

4x3- 

5. 

3 

X*' 

7     x2  +  2  a; 
(x  +  l)2' 
«      1-x2 


(X*  +  1)2 

9.  _L. 

2Vx 

io.  -fx-i 


Page  24.     Art.  8 

(6«-4)6*.       3.    -i(10x-2).       4.    (e—A)  (rf-2) 


Page 

32. 

Art. 

13 

1. 

10X9. 

g 

2  -  6  x  -  x2 

2. 

-  8  x-9. 

(a:2  +  2)2 

3. 

4. 

c 

2Vx 

1 

Vx3 

_  5   4/ — 
4  VX" 

1 

^9. 

10. 
11. 

3x-f  5 

Vx  +  2 

Va(Vx,—  y/a) 

5. 

2  y/x  (  y/x  -fa)  (  Va  +  Vx)3 

6. 

n(x  +  a) 
nx*1-1. 

,n-l. 

12. 

1 

7. 

Vl-a;2(l-x) 

8. 

a2 

3 

13. 

1 

(a2-  x2)s 


2  x(l  -  x2)  +  vT^x2 


355 


356 


ANSWERS 


14. 
15. 
16. 

17. 
18. 
19. 

20. 
21. 


4  oft  +  3  aft 

4A/x^a!  +  ai 

2  x3  -  4  a 

(l-x2)^(l+z2)* 

—  2  nx"-1 
(xn-l)2  ' 

_  m(ft  +  x)+n(g  +  x) 
(a  +  x)™+!  •  (b  +  x)«+! 


x2(x3  +  1)T 
56x3(x2  +  l)i 


6 


dx 


12(m2-  t*  +  l)  — . 
ax 


I.     60M5(1  4-^2)2^?. 

ax 


24.    w  +  x^. 
ox 


du 


25.    (2m  +  6xw)  —  +  3w2  +  4x3. 
dx 

dx  nun 


(a+z)n      (a  +  se)n+1 

27-    2  wx3w  ^  +  w2x3  —  +  3  uWw. 
dx  dx 


30. 


62x 


&x 


Va2 


32.    (0,0),  (i,--il 
\9o        27  a/ 

\9a'      27a/' 
34     (21m3-  19m)10x 

(7  w2  +  5)* 
35.   At  right  angles  at  (3,  ±  6). 


y2  +  2  xy 

2xy +  x2' 


x  +  a 

a 

ax-}-  b 

8x- 

■7 

4x2-7 

x  +  2 

2 

1-X2~; 

4x 

Page  33.     Art.  14 

3     6(2x-3y)2+2(x-y)+l 
9(2  x  -  3  y)*+  2(x  -  y)  -  1 ' 

Page  37.     Art.  18 

7.  ?ixn_1  log  x  +  xM_1. 

8.  nx"-1  log  xm  +  mxn_1. 
x 


1  -x* 
log  x  +  1 


9. 
10. 


x2-l 

1 


2(\/x+l) 


11.  log0e- 

12.  log10e 


12xV2  +  x  —  1 

2  V2+x  (3  x2  -  V2+x) 

2x  +  7 
*x2  +  7x* 


xlogx 

14. 

aeax. 

15. 

4  e4x+5. 

16. 

i 

(i+x)2" 

17. 

ex 

(1  +  e*)2 

18. 

2/  -  3  x2e* 

19. 

1  -  y2. 

ANSWERS 

25. 

21ogx 

26 

1 

xlogx 

27. 

-  (log  x +  1) 

28. 

alogxlPg«. 

X 

357 


e?  +  e~3 


-(x-l)^(7x2+30x-97) 

12(x-2)^(x-3)^ 
2  +  x  -  5  x2 


e1 


2V1-X 


21     1  +  e*  31     1  +  3  x2  -  2  x* 

'    X+e*'  *         (l-x2)f 


.   5  x*(a  +  3  x)2(a  -  2  x) 
(a2  +  2  ax  -  12  x2). 


22.  wx"-1^  +  xnax  log  a. 

23.  V« 

Vx(a  —  x) 

24.  1  33.    («-2«)vY 

x(logx)2'  •  (x-a)* 

Page  41.     Art.  22 


1. 

7  cos  7  x. 

12. 

-  20  x  (3  -  5  x2)  sec2(3-  5  x2 

2. 

—  5  sin  5  x. 

13. 

2  tan  x  sec2  x  —  2  tan  x. 

3. 

2  x  cos  x2. 

14. 

secx. 

4. 

2  cos  2  x  cos  x  —  sin  2  : 

t  sin  x. 

15. 

cot  Vx 

5. 

3  sin2  x  cos  x. 

2Vx 

1                1 

6. 

10  x  09s  5  x2. 

16. 

log  a  •  ax  •  sec2  (ax). 

X2 

7. 

14  sin  7  x  cos  7  x. 

17. 

w  sinw_1  x  sin  (n  +  l)x. 

8. 

sec2  x  (tan2  x—  1). 

18. 

cos2W^. 

9. 

3  sin2  x  cos2  x  —  sin4  x. 

ax 

10. 

sec  x  (tan  x  +  sec  x). 

19. 

mn  sinm_1  nx  •  cos(w  —  n)x 

COSn+l  mx 

11. 

-16x(l-2x2)sin(l 
cos(l-2x2)2. 

-2x2)2 

20. 

2 

1  4-  tan  x 

358  ANSWERS 


21.  cos  (sin  «)  cos  u^  26'    ~  8  csc2  4  *  cot  4  *■ 


dx 
J.   2  ae8*  sin  e8*  •  cos  e8*. 


26.  8  (4x  -  3)  sec  (4x  -  3)2 

tan  (4x  -  3)2. 

27.  -2a?oatfrf+?eoV5tanVS- 


cose^logx+^i-.  2Vx 


28. 


y  cos  xy 


xcosx2  1-xcosxy 

VsTnx2  29.    -  csc2  (x  +  y). 

Page  43.  Art.  23 

4*  6.  1  11.         * 


\/l-4x4  2Vsin-1xVl  — x2  Vl  —  x2 

1  7.         1       •  12.         ^*- 


Vl  -.x2  e*  +  erx  *  Vx2  -  1 

3  3  8.   -1  13.    -ZJL. 

V6x-9x2  xVl-(logx)2  l+.x2 

o  n  SPC2X  . 

4.  *        .  9       /t      t  •  14.   |Vl  +  cscx. 
^1  _  xz                          VI  -  tan2x 

5.  _^_.  10.   — J=-  15.   h 

l  +  x2  vT^T2 

16.  sec2x.tan-ix  +  -^i£L.  17.   sin-ix+- 

1  +  x2 


18. 


gtan    1x  21,     . _ 

1  +  x2 "  2  Vx  (x  +  1) 

2  22. 


19.    —  -  ex  +  e~x 

VI  -  x* 

23 


r 

25. 

L-x2 

2  sin  x 

26. 
27. 

Vl  —  4  cos2 
-  1 

X 

2(1  +  x2) 
-1 

,.       —  2  cos-2  x  +  w2  sin2  x                 V 1  —  x2 

'  "    x2  +  1 '  24.    2.                                 28.   0. 

Page  45.     Exercises  on  Chapter  II 

1.   6x+  15  x2.  4     a2  -  2  x2 

2     -6      15  V^^x2 

x3       x4  5.    log  sin  x  +  x  cot  x. 

3  x  -  1  «          -  a8 


3. 


2Vx-3  *Va2-x2 


7, 


8. 


19. 


1  Ja  +  a 
x  'x  —  a 


ANSWERS  359 

x       V         xj  1— iC4 

2  x2  -  2  x  4-  1 

2(x-x2)2    '  ""'  x 

_      e^  22.  1. 

9>    =  '  cot  «•  ««  „ 

2  Vu  23.  2  tan  «  +  e8ec  x  •  sec  x  tan  x. 

JO.    i-loga.  24.  2^ST. 

x        5  a*  -  y2 

11     -(3x4-  a*)  25         2xy2  +  3x2 

3x2  +  l 


(i  +  x2)i  ' 

,12.   e*(cos  x  -  sin  x).  26.    - 

3  r  +  1 

13- 1^3=^-  27.  y+a^-i. 

2  xy  —  x2 


^Vx^n:  27.  £ 


14. J 9R    4  cos  (21ocrx2-7) 

5  +  3cosx  28     k— ^ l' 

15.  tan-i  J*.  29.    x^  =  2y. 

ya  dx        y 

16.  1  30.   x  =  rnr. 

(-1  +  ^)  32.   x,    y    are  determined  from 

17.  4  tan*  x.  a2y  _  ±  b2x  and  equation 
18        logx                                                             of  curve. 

'    (1-x)2' 


33.    x  =  kit  ±  -. 
4  4 


5  +  3  cos  x  34.   tan-1  2  V2. 

Pages  49,  5Q     Exercises  on  Chapter  III 


1. 

72  x. 

8. 

8  tan  x  sec2  x  (3  sec2  x  - 

-1) 

2. 

0. 

9. 

2  cot  x  esc2  x. 

3. 

3! 

X4' 

5! 

X6  ' 

6  sec4  x  -  4  sec2  x. 

10. 
11. 

12. 

16  sin  x  cos  x. 
24 

4. 
5. 

(1-x)* 
48 

X 

6. 

7. 

e*  logx  +  Ai*—^!. 
x       x2 

2  log  x  +  3. 

13. 
14. 

sinx. 
_  8(e*  -  e-*) 

360 


ANSWERS 


15. 
16. 
17. 
18. 

19. 


21. 


8  z2e2*. 
_4J 
x2' 
aneax. 
(-l)"w! 
(x  -!)"+!' 


\lmx  +  ra^Y 


(-  !)"•  (m+  n-1)! 

(w—  1)  !(a  +  sc)w+" 

m(  -  l)n-i  .  (W  _  1)  ; 

(a  +  x)n 
3jpa 

2/6  ' 


ft4 


24    - 2  a3*y 

(y2-ax)3' 

25.  -yr(*-i)2  +  Q/-n2i 

a:2(y_i)3 


e2*. 


34.    2"-1cos['2a;+  —  V 


3- 

-?/ 

(2- 

-2/)3 

(n- 

-1)! 

2. 

and  1 

3. 


r,  max.  ;  . — ,  min. 

V3  V3 


Page  53.     Art.  28 

Inc.  from  -co   to  i  ;  dec.  from  1  to  1;  inc.  from  1  to  +  oo  ;  | 

Two.    +  1  at  x  =  \  ±  V^  ;  -  1  at  x  =  J  ±  VJ.       4.    ±  tan-*  ^. 

Page  60.     Art    34 

6.  —  1,  max. ;  —  £,  min. 

7.  —  2,  min.  ;  1,  max. 
2,  max.  ;  3,  min.                               8.   e,  max. 

2,  min.  ;  f,  max.  9.   2n7r,  min.  ;   also  tan-i  ±  V| 

(2  w+J)w,  max. ;  (2ra  +  £)7r,  for  angles   in   2d    and   3d 

min.  for  all  integral  values  quarter.  (2  n  +  1)  71- , 

of  w.  tan-1  ±  V|,    1st    and    4th 

quarter,  max. 
10.    2,  min. ;  —  1,  max. 

11.    x=  3m  +  4      m2  +  44w  +  4=0. 
2(u-l)' 


-,  mm. 
4 


Pages  63-67.     Exercises  on  Chapter  IV 

Two  thirds  the  length  of  the  segment. 

The  parts  are  equal. 

h  5.    -*=•  7.     2r 


V3 
6.   3  inches. 


V3 

8.   Area  is  ~ 

2 


ANSWERS  361 

9.  The  side  parallel  to  the  wall  is  double  each  of  the  others. 

10.  The  altitude  is  equal  to  the  diameter  of  the  base. 

11.  8  inches. 

12.  One  mile  from  stopping  point. 

13.— Most  economical  per  hour  at  15  knots. 

14.  fa. 

15.  The  altitude  of  the  rectangle  is  equal  to  the  radius. 

16.  The  altitude  is  equal  to  the  radius  of  the  base. 

20 

17.  —  -  yards  from  the  nearest  point. 

>         42- 122 

19.  15  V2  feet. 

20.  The  diameter  of  the  sphere  equals  the  edge  of  the  cube. 

21.  ilfeet. 
V2 

22.  Circular  arc  is  double  the  radius. 

23.  — ,  T>  being  the  distance  between  the  centers  of  the  spheres. 

24.  Arc  =  2  *r(l  —  Vf) . 

25.  Angle  at  center  of  variable  circle  denned  by  d  —  cot  6. 

26.  The  line  should  be  bisected  at  the  given  point. 

27.  The  altitude  is  f  the  slant  height  of  the  cone. 

28.  (a* +  &*>*.  31-   x  =  a'P-  34.   *=J2jD?feet 

29.  |  a.  32.    20  ft.  X    » 

35.    tan  6  =  sec  0— tan  </>• 

30.  a +  6.  33.   aV\.  36.    6  =  35°  20'. 


Pages  76,  77.     Art.  39 

3.  About  3°  58'  per  second.  5.    (3,  -\e). 

4.  120  feet  per  minute.  6.    At  5a/2  miles  per  hour. 

7.  (3,6).  9.    2ab.  11.   5  w. 

8.  At  60°.  10.    i  2.  12.  2. 


362  ANSWERS 


13. 

1  and  5. 

16. 

S~64' 

'~32 

17. 

a 
16' 

Page 

19.  ±  16,  =F  12  feet  per  second. 

20.  sin  0  •  d<t>. 

36 

21.  — :rrL  radians  per  second. 

V37* 


Page  83.     Exercises  on  Chapter  VI 

1.   -J^+JE,  2V^x,  4tt\/^+  ax,  4wax.  2.    2,    2. 

*    x  y    x 

3.   secx.  5.    ,-^(a2-a:2).  7.    V2  ap. 


az 


4.   Tra2^  6    pVl  +  (loga)2.  p 

9.    30  7t,  72  ir.  10.    f  sin  d,  90°,  270°  ;  2,  -  2. 

Pages  87-89.     Art.  48 

I    Xix  .  yiy  _  1  4.    (a)  x  +  2  */  =  4  a, 

a2        62         '  y-2x  +  3a  =  0. 

y      yi      Mb/  y  =  T2x±3. 

2.  y  =  x.  (7)  */  =  x+p,  x  +  y-3i>=0. 

3.  2y  =  9x  -3,  9y  +  2x  =  29.  5.    3.  6.    4^17. 
7.    (a)  Parallel  at  points  of  intersection  with  ax  +  hy  =  0. 

Perpendicular  at  points  of  intersection  with  hx  +  &?/  =  0. 

(j8)  Parallel  at  f—^-z,   8  *y  2>) ;  perpendicular  at  X  =  0. 

(7)  Parallel  at  (—,   fL^iiA  .  perpendicular  at  (0,  0);  (2  a,  0). 

8. _  = ,  i.e.  they  must  be  confocal. 

a      b      a'      b' 


IT 


12. 


2  2  nx 

13.    2c  +  a.  19.    (2p,  ±2pV2). 

o 

Page  95.     Art.  51 

1.    An  inflexion  at  x  =  y  =  2. 

,     /2o     3a\      /-2o     3o\ 


ANSWERS 


363 


8.    Point  of  inflexion  at  (a,  $a),  tangent  is  x  +  y=-£.     Bending 

o 

changes  from  negative  to  positive. 

10.    (-1,l),(2±V3,^^g). 


Page  103.     Art.  57 

1.  y  =  0,  x  =  a,  x  =  —  a. 

2.  x  =  0,  x  =  2  a,  2/  =  a,  y  =  —  a. 

3.  y  =  a,  y=—a  ;  two  imaginary. 

4.  y  =  « ;  x  =  c  twice. 


5.  y  =  —  x  +  -  ;  two  imaginary. 

o 

6.  x  =  1 ;  one  parabolic  branch. 


8.  a*  =  0  twice  ;     one    parabolic 
branch. 

9.  x  -  0,  y  -  0,  x  +  y  =  0. 

10.  y  =  #  ;  two  imaginary. 

11.  x  +  y  +  a=0;  two  imaginary. 

12.  y  -f  sr,  =  0  ;  two  imaginary. 

13.  x  =  0  twice  ;  x  =  y,  x  =  —  y. 

14.  y=x,  y=—  x;  two  imaginary. 


7.-  x  =  —  a,  y  =  —  6,  y=x  +  b-a.        15.   a;  +  2y-0,x+y=l,a--y  =  —  1. 


Page  107.     Art.  60 

1.  yff  =  6. 

2.  Polar  subtangent  =  — ,  Polar  normal  =  y/a?+p\  Polar  subnormal  =  a. 

3.  yff  =  -  +  2  0,  Subtangent  =  -  p  cot  2  0,  Tangent  =      q2p     . 

2  Va4-V 

Subnormal  =-a'2sin2*,  Normal  =  £ 
P  P 

5.   ^,2asin2^.tan^. 
2  2  2 

7.   They  have  a  common  tangent  at  the  pole  ;  elsewhere,  -. 


3.    1. 


Page  111.     Art.  62 

4.    (ar  +  y)  cos  xy. 


5.    1. 


Page  115.     Art  63 
5.    \  square  units.  6.    5Vlo! 


Differs  by  dxdy. 


364  ANSWERS 

Page  117      Art.   65 

3     _  ax  +  hy  +  g  .  ?  3 

hx  +  by  +f  '    VlT^' 

4.    x~.  8.    *=1. 

y3  &>  +  x 

6.    2l  +  Vy+U.  9.    yCco8(gy)-e^-2or1> 

s  s  x  [x  +  e2*  —  cos  {xy)~\ 

Page  121.     Art.  66 

2.  8  a;  +  8  y  -  z  -  12  =  0.  5.   —  +  JL  +  JL  =  at • 

£Ci*      ^      z\^ 

3.  »=Js=t^=«=»  7.   008-i_i<L.         8.    JL. 

1  -4  3  V119  V17 

9.  2*1*  +  „»-«1.  =  0;  ^fl  =  r=H=st^l. 

2  5Ci  «/i  —  *i 

10.   2x  +  2V3y  +  3s  =  25,        x  +  s  =  5. 

Pages  128-130.     Art.  72 

2.    **-2y^  =  0.  11.    ^  +  ^+y=0. 

dy'2  dy  dy2     dy 


Q 

r    h(f)T 

d2x 
dy2 

4. 

fd%x      cPx\  dx_Q 
\dys      dy2)  dy 

5. 

£*  =  cos's +  2(-T 

dx2                       \dxJ 

6. 

dy* 

8. 

^L  +  y  =  0. 

du2 

9. 

SUO. 

12.   2z 


d2z  ,  n  f dz 


+  2(^\ 


dx2         \dxj 


+  (l-z2)2z  —  +z*  =  0. 
dx 

13.  ff+«/=0. 
dt2 

14.  ^  +  ^  =  0. 

d£3 

d«2       <  d« 


16. 


dl>  17.    -6. 

Page  137.     Art.  74 

6.    Divergent.         7     Convergent.        8.    Convergent  in  both  cases. 


ANSWERS  365 

Page  140.     Art.  75 

6.    -1<x<1.  7.    |s|>l.  8.    -a<x<a. 

Page  145.     Art.   77 

2.  '         /(s)  =  (x-l)3+(x-l)2  +  4(a:-l)-3. 

/(1.02)  =  -  2.919592,  /(1.01)  =  -  2.959899. 
/(.99)  =  -  3.038901,  /( .98)  =  -  3.079608. 

3.  3(y-3)2+4(y-3)-8.  4.   sin 31°  =  .51508. 

Page  149.    Art.  78 

1.   x+??+2-x*+B.  3.    .000002. 

3      15 

6.    .017452.  7.    1  — —  —  — +  ^. 

2       8 


y/S     l/„      tt\      V3 

2        2 


9.    ?  +  e?K  + —h*  +  B. 
10.    15  +  24 (a:  -  2)  +  13  (x  -  2)2  +  3  (a  -  2)8. 

U.log«  +  *-  ".  +  -»-».+*. 

8    .     a;      2  a;2     3  Xs     4  a:4 

12.    -  4  (a;  +  1)  +  6  (x  +  l)2  -  4  (x  +  1)3  +  (x  +  l)4. 

14.  5.013.  16.   3.433987. 

15.  11.0087.  17.    .0127.-.. 

18.   1  -  (*  -  1)  +  (x  -  l)2  -  (x  -  \f  +  B.     0  to  2. 


Pages  159,  160.     Art.  83 


2. 

2  a2 
a2  +  62 

3. 

13 

7  ' 

4.  JL. 

2a 

Pages  163, 

164. 

Art.  85 

3. 

f 

5. 

f 

8.    -4, 

4. 

4. 

7. 

*■ 

366  ANSWERS 

Page  164.    Art.  85 


1.    0. 


6 


9.   1. 


2-  2.  6.    I.  10.    _?, 

n  3 

3-  3.  7.    1.  11.    I 

]oga 
log  b ' 


4.   !^?.  *    1. 


Page  166.  Art.  86 

1-    1-                                          3.    0.  5.    -$. 

2.    0.                                          4.    5.  6.    1. 

«•     e  2c2. 

Page  171.  Art.  89 

1.  First.  6.    Second. 

2.  They  do  not  touch.  7.   a  =  —  1. 

3.  Third.  8.    y  =  2  a;2  -  5  x  +  4. 

4.  3i(x-a)  =  a(2/-a).  9  a.    (—  2,  -  8),  First. 

5.  y  +  12x  =  10.  96.    (- 2,  -8),  Second. 
10.   y=-x2  +  2x  +  3,  $x=-3y2+  Uy-36. 


2. 

First. 

13.   Second. 

2  a. 

a . 
2* 

Page  179.     Art.  97 

6    (*2  +  rcV)2 
n(n  —  l)xy 

i. 
2. 

9    aVx(8a-3x)* 
3(2  a  -  xy 

3. 
4. 

00 . 

(z2  +  ?/2)* 
2  m2 

7     2(z  +  2/)^ 
Va 

10.  «!. 

a 

5. 

(e2x2  -  a2)^ 
a& 

8.   3(aa*/)*- 
Page  181.     Art.  98 

3     a(5- 
9- 

11.  ( *    _j_y 

V3V2    54V2"/ 

1. 
2. 

pVl  +  (log  a)2- 

-  4  cos  0)* 

-  6  cos  0 

3, 


ANSWERS  367 


2p*  6     4x/^ 

3 


Va 

5.     -if. 

a2 

1.   a  =  0,  /3  =  0. 


7     q(l  +  gg)* 


Page  188.     Art.  100 


2.    a_alog-— -  ,  p--.      b     («  +  i8)!-(«-/S)i  =  (4a)t. 


.    a=x-^(ea-e   a),p  =  2y. 

.   (a  +  )3)f-(a-/3)§  =  (4a; 

.   ,  1K    .  ,   .        6.    (aa)*- (6/3)'- =  (a2  +  62)t. 

3     ce  =  q  +  15  y      8  =  ay  ~  °  y 

6  a-V  2  a*  7.    (a  +  B)%  +  (a  -  /3)t  =  2  ai 

16.    a  =  a(0'  -  sin  6"),  p  =  a(l  -  cos  #'),  0'  =  0  -  tt. 

Pages  198,  199.     Exercises  on  Chapter  XIII 

1.  (0,  0)  ;  ax±by  =  0.  6.    Two  nodes   at  infinity  ;   the 

asymptotes  arex  =  v-|-l,  x+y=±\. 

2.  (O,0);cuspoffirstkind,y=0.  m     ,n         x     ,  '      /       _ 
V   '           .                            *"                  7.    (0,  -a);   (+a,  0);   (-«,0:) 

3.  Four    cusps    of    first    kind ;    the  tangents  are,  respectively, 
(0,  ±  a),  (±  a,  0)  ;  y  =  0,a;  =  0.  V%  +  a)  =  ±V2a; 

2(.z+a)  =  ±\/3y; 

4.  (0,  0)  ;  conjugate  point  with  2(x  -a)  =  ±VSy. 

real  coincident  tangents,  y  =  0.  0     /  Ax 

°        '  "  8.    (—  a,  0)  ;  conjugate  points. 

5.  (0,  a);    y  =  a  +  *J    cusp   of  9     (0,  0)  ;  »  =  0,  y  =0. 
second  kind.                                                10.    (0,  0)  ;  is  a  tacnode  ;  y  =  0. 

12.    Terminating  point  at  (0,  0) . 

Pages  207,  208.     Exercises  on  Chapter  XIV 

1.  x2  +  yl  =  p2.  7.   y2  =  4  a(2  a  -  x). 

2.  xf  +  i/f  =  ai  »•   &2x2+(a2fW=62(a2  +  &2). 

3.  x$  +  yl  =  cf.  9-  (*2  +  ?/2)2  =  4  c2(^2  "  y'2)' 

4.  4xy  =  k*.  10-  16y3  + 27^  =  0. 

5.  (x  -  a)2  +  (y  -  py  =  r2.  11-  y±s±fc  =  0. 

6.  y2(x  +  2a)  4-a*  =  0.  12.  (x2+y2-ay)2=a2(x2+0/+a)2). 

13.    62x2  +  a2y2  =  4  a262. 


INTEGRAL   CALCULUS 
Pages  215-216.    Art.  114 


1. 

„      3 

21. 

—  cos  nx. 

2. 

xa+l 

n 

a+  1 

22. 

x  .  sin  2  x 
2          4 

3. 

4. 

2  to     X«H 

am~\L      ,. 

23. 
24. 

x     sin  2  x 
2          4 

cos  (to  4  n)  x 

5. 

ax  -  \  aH*  +  |  a*05*  -  £  x2. 

m+n 

6. 

5kJx-f -J* JL. 

e         2x2     3x8 

25. 
26. 

—  |  COS  X2. 

sin  x  —  ^  sin3  x. 

7. 

H^2  +  «2)3- 

27. 

—  cos  x  +  I  COS3  X. 

8. 

(atx  +  &)n+1 

28. 

tan  x  —  x. 

JoT+1) 

29. 

|  tan3  x. 

9. 
10. 

log;(x  4  a). 

|  log  (2  ax  -  x2). 

30. 

—  cot  (ax  4  6). 
a 

11. 

—  log  cot  X. 

31. 

-  f  (cot  x)l 

12. 

—  log  (1  +  cos  a). 

32. 

log  tan  x. 

13. 

log  (logx). 

33. 

\  sec3  x. 

14. 

flog(x34l). 

34. 

—  cos  X. 

15. 
16. 

—  log  COS  X. 
log  sin  x. 

35. 

sin-i^. 
a 

17. 

-  eax. 

36. 

i  sin-12a;. 

18. 

a 

37. 

a           a 

19. 

(a  4-  b)m+nx 
w  log  (a  4- 6) 

38. 

A  tan-i^- 

ab             b 

20. 

|  sin  2  x. 

39. 

tan-1  (x-2). 

Page 

219. 

Art.  115 

1. 

x  sin^x  +  Vl  —  x2. 

3.   _X*  Sill  x  -f-  Z  X  COS  X  - 

2. 

ex  tan-1  ex  —  \  log  (1  4-  e 

*). 

4.   J2±i(log*         * 
n  +  1  V               n  +  1 

ANSWERS  369 

5.  A[2z3tan-ix-z2+log(l+*2)].     Q    £e*(sin  x  +  cos  x). 

6.  sec  x  [log  cos  x  +  1J.  10.    |  e*  (sin  x  -  cos  x) . 

7.  J  [(*» +1)001-1*  +  *;].  11.    icosxsin2x-icos2xsinx. 

8.  Ksin3*-3*cos3*].  12.    x  tan  x  +  log  cos  x. 

Pages  220-222.     Art.  116 

4.    Ksin-i*)2.  i,     ,      ♦      « 

11.  log  tan-. 
6.    Jcos(*2+l)[l-logcos(*2+i)].  2 

12.  log  tan  g  +  |). 

13.  -§(a3-x*)i 


7. 

sin-i  « 
a 

8. 

I  tan-i  « 
a            a 

9. 

±cos-i«. 

a           * 

10. 

sin-i^-^ 
a 

14.  log  (x  -  1)  - 

15.  1 


2 


x-1      2(x-l) 


2  sin2  * 


16.    -^tan-^v^tan*). 

V2 


Page  226.  Art.  118 

1.  -i-logV^Cg+l)-!.  3.    _±lo|f*=*. 

2V2         V2(x+1)+1  12     °x+l 

2.  -i_  tan-ii^l.  4.   £sin-i(3*  -  5). 
Vl4             V14  ' 


5.  Vx2  +  2  x  +  2  -  log(x  +  1  +  Vx2  +  2  *  +  2). 

6.  - V-*2  +  2 x  +  1  +  sin-i^ni. 

V2 

7.  .-V/8-4*-4*2+3sin-i2*_+l- 

2  3 

8    ^V3x^  +  x-'2-~  log(aj  +  |  +  Vx2  +  i  x  -  f ). 

9-   ^Vl+*-2*2  +  -JL_sin-i43-lt 
2  4V2  3 

10.  ^ log  (x- 5) -| log  (x-1). 

11.  Vax  -  x2  +  ?  sin-i  2-^J? . 

2  a 

el.  calc. — 24 


370  ANSWERS 

12.    -  V-  2  x2  -  3  x  -  1 - 


sin"1  (4x  +  3). 


13.    -lVi-2*-3^--^-sin-i3a;  +  1 
3 


2  V2 

10 

3V3 


ilog(«_±^) 


Page  227.     Art.  119 
2 


3.    -lo 

4. 


l-2x  +  V5^-4a;+  1 


/a+V«2-a2\ 


1    ,       /V2  +  Vx2  +  2z  +  3\ 
~VSlQg(  "¥+1  J 


lo< 


/ 1  — a;  +  2  Vz2 +  x+J  \ 


6.   sin- 


V2  (x  -  1) 


7.    _l]o(f/l-x+V-x*-10x-r 


9. 


a2z 


10 


5C  +  2 

Vx2  +  a'2 


—  Va2  —  x2 


11.    -JLrin-i      3~2*_ 

V2  V3(2  x  -  1) 


Pages  227-228.     Exercises  on  Chapter  I 

3.    i  log  (6  x^  +  12  a;  +  5) 


2.    5  tan-i-. 

8  2 


4.   *£(!  +  **).  5.   -f(3-2, )t      6.  ^ 


7.    _|(d«_rf) 

10.    I  sin  8  x. 
12.    —  cos  e*. 


V,2  +  1 


Ks  +  l)1- *(*-!)*. 


11.  1  log  tan  ^^  +  -1 

3     s        V  2       4J 


13. log  (a  cos  se  +  b). 

a 


14.    -  log  (e~x  +  Ve~2x  -  1) . 

15.    I  sin-i  x2. 

2V2        L  *  J  2     &e*  +  l 

18.   i  a*  tan-i  x  -  ^  *4  +  rV  *2  -  tV  log  (*2  +  1)  ■ 


ANSWERS  3T1 

19.    -f2  +  2xloga  +  (x\ogaf}^       2Q    tiind_secd. 
a* (log  a)3 

.21.    -cot?.  22.    logCgcoB'g  +  ftBin'^ 

2  2(6 -a) 

23.    -JVl-logx.        24.    ilog(e'^  +  l).        25.    sin-i / 2 sin ^  +  1 V 
26.    tan-1  (logic).  27 


6(a  —  6  tana;) 

2' 


28.    irin-ir^-^+^1 

L       V5  a2      J 


V3x2  +  2x+l  +  log  p  +  l+V3*»  +  2s+r 


30.    -  —  log  (a;  +  Vx2  -  a2)  + 


Vx2  —  a2 


2  x2  2  a2x 

31.    —  cos  x  log  tan  x  +  log  tan  - .  32.    \x  —  \  log  (sin  x  +  cosx). 


Pages  236-237.     Exercises  on  Chapter  II 

1.  *(s  a-2  _  2  x2)  Va2  -  x2  +  —  sin-i?  . 

2.  ir_JL_  +  ltan-i2l.  5.   -  Va2  -  x2  +  ^sin-i?- 
8|_x2  +  4      2           2j  2  2  a 

3.  2a;-1       -f-i-tan-i^l.     6.    -  ^ZEZ. 


3(x2-x  +  l)      3V3  V3 


4.    _?v/a2-x2+—  sin-i-.  7. +     '   2x_ 

8.    ~  (2  x2  +  5  a2)  Vx2  +  «"  +  —  log  (x  +  Vx2  +  a2) . 


9.    -  Vx2  +  a  +  ^log(x  -f  Vx2  +  a). 

10.   -(2  x2  -  ax  -  3  a2) V2  ax  -  x2  +  ^  sin-i  — -  • 
6  2  a 


n         (2ax-x2)?  12>    _Vl 


3  ax3  2  x*< 

13     3fa  +  2)«-5(g  +  2)       3t     x  +  1 
8(x2  +  4  x  +  3)2  16     °  x  +  3 


372  ANSWERS 

14.    l(x  +  l)Vl-2x-x2  +  sin-i «±i 
2  V2 

V^EI-lrin-ll.  18- 


17. 


SILL •  / - r- 

2  x2  2  x  aVa  +  bx2 

-1  19.  -cos* 


2(x2  +  7)'  (1  +  e)  Vl  +  esin20 

20.  -(2x2  -  a2)Va2  -  x2  +  ^siiT^- 
8  8  a 

21.  —  (33  a4  -  26  a2x2  +  8  x*)  Va2  -  x2  +  ^-  sin-i  -  • 
48  16  a 

Page  241.     Art.  122 

1.  JLlog^.  3.    ^-4,+^logi^I3 

2c       x  +  a  2  T  2      °    x  +  1 

2.  w  (x  +  1)2  ■  4-    x  +  log(x-a)«(x-&)». 

x(x  —  1) 

5.    2_±^§log(x-2-V3)-^^log(x-2+V3). 
2V3  2V3 

6    li0(y(^-l)^-2).  7.   i og  («-«)(« -6>. 

20    &  (2x+l)(x  +  2)  x-c 

8.  x  +  ^—  [a2log(x  +  a)-&2log(x  +  6)]. 

&  —  a 

9.  log[(x  +  2)V2x-l].  12.    ^-7x  +  641og(x  +  4) 

10.log(x-a)(,  +  6J,  -271og(x  +  3). 

2  a&        ax  +  & 
t1      1.  x6 14     I  iog  !  +  ^ . 

u-  3l0g(2-T^xr^^  2logi-. 

Page  243.     Art.  123 

L    "2(x-l)+Il0g^l*  3"    2(a2  -  x2) ' 

2-  ^  +  log^  4'  v^Tvii* 

5.    ^_2x  +  ^^  +  logx(x4-l)2. 

2  x2  +  x 


ANSWERS  373 


6.    log(x2-a2)--^a2 


x*-a*  4(V2  +  l-a;)2 


8.    ax--  +  log 


x  x  +  a 

9.    x  +-L-l[28\og(x  +  S)-\ogxl 
ox     y 

10.    Aiog^±_^_±.  ll.   a;__A^_51og(x-3). 

a2  x  ax  x  —  S 


Page  244.  Art.  124 

1.    log *— .  2.    llog  aj,  +  1a  +  -tan-ig. 

°V^T4  4     *>  +  l)2^2 

3.  —  [log(x  +  a)--log(x*-ax+a2)  +  V3  tan-* 2 *  ~_a~| . 
3a2L                         2  aV3  J 

4.  -  1  tan-*  -  +  -  tan-i  ? .  7. —  tan-i* 

a           aft           b  2  a  (x  —  a)      2  a2           a 

6.    __Ltan-i-^5— .  9.   log    x  ~  1    -tan-*  a. 

V3            2x2  +  l  V^+T 


Page  24b.     Art.  125 

1.    tan-i*  +  _^-.  2.   Atan-i«U?72<* 

x2  +  1  2  a  a      2  (x2  +  a2) 

3.    llog  *2  +  1    4-     z-1     . 

4     &(x  +  l)2      2(z2+l) 

4    -g  +  Slog^^-    3  +  2*    -Stan-ix. 
x  x  2(.r2  +  l) 

5. +  log  (x2  +  a2)  -  J_tan-i?# 

2  (x2  +  a2)  5  V  J      2  a  a 

6.  — - +ilog(x2  +  l). 

x*+l      4(x2  +  l)2     2     &K  ' 


374  ANSWERS 


Page  249.     Art.  127 


1.   log  Vx  +  1  ~i  .  2.   2Vx-3?/x  +  Qfyx-6log(fyx  +  l). 

y/x  +  1  +  1 

3.    2  log  ( Vx  -  1  +  1)  + 


Vx  -  1  +  1 


4.   2tan-iVx-2.  5.    llog^      a  ^-& 

&       Vx  —  a  -f  6 

6.    14  (xT?  -  i -xt  +  J  xt?  -  |  x7  +  i  XT?). 

Page  252.     Art.  128 


L         »l-xj      x-l  +  Vl-a:2 
2.    _21og[-V2  +  V^]. 

Pages  253-254.     Exercises  on  Chapter  IV 


1.  2  V3  tan-i  V^V^  ~  V^  ~  1# 

.    3  l        3  V2(x-a)*-l 

2.  o(x-a)^ -log *> '- . 

*  4V2        V2(x-a)^  +  l 


3.    x-4Vx+ 1 +  81og(Vx  +  l +2). 

2  2Vx^T  +  l 


4.  log(x+Vx-l) -tan"1 

V3  V3 

5.  M2x-3a)(a  +  x)i  ?         1      1q„  V^IT^r  +  sVS 

6.  6  log  (**  -  3xU  5).  "    2^«2    °  V*2  -  ^  "  *V* 


8.  i[x2-xVx2  -  1  +log(x+  Vx2-1)]. 

9.  f  x^  -  f  x^  +  f  x*  +  2  x^  -  3  x*  -  6  x^  +  3  log  (x*  +  1)  +  6  tan"1!*;*. 

10.  -$(2a*  +  s2)V^V.  vgr^ 

,,        . a,  „  Va2— x'2— a  a2x 

11.  Va2-x2  +  ol0S 


2        Va2-x24-a 


Va2  -  x'2       •   _,  x 


14. 


12.    -  — ^_sin-i-.  a2Vx2  +  a2 


ANSWERS  375 


15     x(2x2  +  3a2)_  lg     1^    Va*+z*+x      Va2  + 

3a4(x2  +  «2)^  2     °  Va^+x^-x  x 


17. 


«Va2  -  x2 


Page  256.     Art.  131 

5.  f  esc3  x  —  cot  x  —  |  cot3  X. 

6.  -  64  [cot  4  x  +  -|-  cot3  4  a] . 

3.  tan  x  +  |  tan3  a;  +  A  tan5  x.  1 

3  7. —  +  log  tan  x. 

4.  —  128  [cot  2x  +  cot3  2  a;  2  tan-  a; 

+  |cot52x+ 1  cot72x].  8.    -  \  cot3  x  -  i  cot5  x. 

Page  257.     Art.  132 


1. 

£  sec4  £  —  \  sec2  x. 

5. 

\  sec4  x  —  sec2  x  + 

2. 

—  4.  esc7  x  +  |  esc6  x 

—  |  esc3  a:. 

6. 

sec"-1  x     secn-3x 
n  —  1          w  —  3 

3. 

-  [  -sec5  ax — sec3  ax  +  sec  ax ) . 
a\5                3                          / 

7. 

log  sec  x. 

4. 

—  (sinx  +  cscx). 

8. 

—  log  CSC  X. 

Pages  259-260.     Art.  133 

1.    _iCot3x  +  cotx  +  x.  tan-ix 

4. 


2.  —  tan2  ax log  sec  ax.  w 

2a  a     ° 

3.  i  (tan2  x  +  cot2  x)  5-    i tan?  K  -  I  tan5  «  +  i  tan3  « 
+  4  log  (sin  x  cos  x) .  —  t'an  x  +  x- 

Page  260.  Art.  134  (a) 

1.  —  COS  X  +  |  COS3  X.  _4                         2                         8 

_         .        ,  4.    |  cos   3  x  +  3  cos3  x  —  |  cos'  x. 

2.  —  |cos5x  +  \  cos'  x.  8 

3.  log  sin  x  -  sin2  x  +  J  sin4  x.  5.    |(1  -  cosx)z  -  |(1  -  cosx)*. 

Page  261.  Art.  134  (6) 

1.  —  $cot3x.  

2.  -cotx-fcot3x-icot5x.  5-    !Vtanx(tanx-3cotx). 

3.  _  cot5  x  (i  +  \  cot2  x).  6.   tann"ly  +  taD"+lx- 

w  —  1  w  +  1 

4.  —  I  cot3  x  —  2  cot  x  +  tan  x. 


376  ANSWERS 

Page  262.     Art.  134  (c) 

1.    |  x  —  ■£%  sin  4  x. 

2-    xis (^  #  +  t  sin3  2  £  —  sin  4  x  —  \  sin  8  x) . 
3.    T|s(3x  —  sin  4  a;  +  fsin8x). 
4.    |(3  x  +  sin  4  a  +  |  sin  8x).  5.    tan  x  +  £  sin  2  x  —  §  x. 

Page  264.     Art.  134   (d) 

1.    |(x  —  sin  x  cos x) .  2.    ^  cot  x(cos2x  —  3)— §x. 

3.  —  |  sin3  x  —  sin  x  +  log  tan  (-  +  -]. 

4.  —  \  cot  2  x  esc  2  x  +  J  log  tan  x. 

5.  |  x  +  2  cot  x  +  \  sin  x  cos  x  —  £  cot3  x. 

Page  265.     Art.  135 

1.    Jtan-i(^tanx).  2.   -  tan-i  f~2  tan  (*■  -  -\~|. 

1     ,      tanx-2-V3 


3. 


2  V3        tan  x  -  2  +  V3 


x 


b  tan  -  -  o  +  vV  +  b2 

4.   — —  log . 

v  a2  +  62       6  tan  ^  -  a  -  Va2  +  52 

-1  7.   J-tan-if^Y 

V2  W2  / 


a  (a  tan  x  +  &) 


6.    i-tan-if^i^y  8'   2 

a&  \      b      J 


li  2 

log 


tan  -  -  3 


2 


Pages  266-267.     Exercises  on  Chapter  V 

3.   2  Vtan  x.  .  4.    J  tan4  x  +  \  tan2  x. 

5.    -  cscx  +  log  tan  I- •  +  -  j. 

6.    -  tan2  x  sin  x  +  §  fsin  x  -  log  tan  (*  +  *\~]  .  7.   ^  +  x  •   • 


ANSWERS  377 

?,  x  11.    I  ex(sin  x  +  cos  a:  —  f  sin3x 

8.  62^sin-  +  cos-j.  _;Cos3x). 

9.  —  $  e_I(sin  2  a;  +  2  cos  2  x).  15     sin^x  _  sin"+3x 
10.    i  e2x(2  -  sin  2  x  -  cos  2  x).  w  +  X         n  +  3 

16.  §tan*x  — 2Vcotx. 

17.  -  32  cot  2  x(l  +  1  cot/2  2  x  +  $  cot4  2  x). 

18.  J-  tan  |  x(l  +  |  tan2  \  x  +  \  tan4 §  x). 

19.  log  tan  2  x.  20.    -  8  [cot  2  x  +  ^  cot3  2  x] . 


21  1    /Va2-x2\« 

22  ^^  +  ^6^^^,  +  ^^ 

-•  i(S+|)-+i(5-|)ta«^+^>- 

24.  x  cos  («  —  &)  +  sin  (a  —  b)  log  sin  (x  +  6). 

Pages  281-283.     Art.  142 

2.   fp*A  3.    2.  4.    36 V3.  5.    40V5. 

6.    f.  7.    fv^.  8.    i.  9.    £.  10.    Trr2.  11.    f|. 

12.   4  a2  tan-1  ^i- ;    4  Tra2.  13.    *•«&.  14.    if*  -  72  log  2. 

2a'  2 


15. 

*.-<*              16     24-8^            17     f            lg<    ^      fl)| 
log  a                                 5 

20. 

3  2                 OJ        *  .           ^      -|-  I 

To3'            '    2'  2(6'-  1)  " 

22.    a2logra.        23.    1.        25.    -. 

&                                       6 

26. 

A.      27.    4a&tan-15.       28. 
T5                                a 

2V2.     29.   I°£i.     30.    * .  IzlT" 

7T                            7T              7T 

31.    ^(4-tt). 

2  V  ; 


Pages  287-289.     Art.  143 

Ex.  4.    2-.  1.    4.  2.    oo.  3.   3.  4.    4  a2. 

2 

5.    oo.  6.    3tt«2.  7.    2tt. 


378  ANSWERS 

Pages  290-291.     Art.  144 

1.   vab.  2.    lira2.  3.   f. 

4    lZ.  5    3y(g2-68)a| 

4  8a6 

Page  292.     Art.   145 


2. 

a\ 

7. 

2 

n. 

h 

3. 

4. 

Ti-r2. 
3  7rr2# 

8. 

«27r3 
24 

12. 

!• 

5. 

2 

25  7T. 

9. 

-log 
2     * 

ft 

;)• 

13. 

3tt 

8   ' 

6. 

c  / 
2(P1" 

-/>2). 

10. 

n 

14. 

7T5 

5  ' 

Pages  296-297.     Art.  148 

In  the  following  answers  the  values  are  given  for  Simpson's  formula 
only,  unless  the  trapezoidal  formula  is  called  for  in  the  problem. 

1.  22  ;  21.5.  3.  0.500014  ;  0.500002  ;  0.5000014  ;  0.5000011. 

4.  5.2523.  5.  37.8555;  36.5261.         6.    0.9996.           7.   8.0047. 

8.  39.6465.  9.  0.7593.                         10.    0.7468.         11.   0.4443. 

12.  335.  13.  3.006.                            14.    1.1873;  1.1830;    11931. 

15.  0.5633. 


6.  \irab. 

7.  \irab. 

8.  \AK. 


Pages  300-302. 

Art. 

149 

a     irabp 
9.    _. 

10.    4£  cu.  ft. 

12. 

|  wa8  cos4  6. 

13. 

1  trabc*. 

11.   fa3. 

14. 

fr8. 

Pages  304-306.     Art.  150 


1.    jrL(io_3  7r).     2.    27rrl(2r2+a2)Vr2-^-ar2sin-i^2^1. 
6V2  L3  «       J 


3.    7tt.  4.    -7T 

3 


|«i«  +  ax{*  +  4  eflxt  +  8  a3  ]og  (~=^\  ] 


ANSWERS  379 


5.    2  7r2a3.  6.    i*^.  7.    firr3.  8.    x/fc3  ;  oo. 

o 

9.    ^^-  10.    Sf.  11.    4  «■*««. 

105  15 

12.  7rr8a3log^-4a2(2a-?/i)];  oo.       13.   5*-2a3.       14.    ^~- 


5.  7rr8a3log^-4a2(2a-yi)l;  oo.       13.   57r2a3.       14. 


Pages  308-309.     Art.  151 

1.  p[V2  4-log(l+  V'2)].                          6la  3    logVS 

'216'  •       5       • 

4.    6  a.                      5.    2irr.                    6.   -  (e  -  e"1).  7.    }f- 

8.    2-V2  +  log1i^.                        9.    4(«3-&3>.  10.   alogS-^a. 

V3                                        aft  &       * 

Page  310.     Art.  152 


1.   (p2-pi)Va2  +  1.  2.    2ira.  3.  8  a. 

4.  a  [tan*  sec  I  +  log  tan  (1+^~|*2. 

5.  ^[flV^  +  l  +  log  (5  +  V^  +  l)]  \ 

c     (7r2  +  4)t      8  w    A   ,     2  ,_ 

••    — f-     -3  7.    4+  — log(V3+2). 

9.    2a[~V5-2-v/31og    V5+V5.T 
L  \/2(2  +  V3)J 


8.   3ira  a    o-f,/5      o      ,/5u„    V3  +  V5  1 


2 


Page  311     Art.  153 

1.    8a.  2.    JJWL.  3.    6a.  4.   la*!2. 

ra  —  1 

7.    ^[(4  +  9ti«)*-8]. 

Pages  313-315.     Art.  154 
1.       Tra^l-^V  3.    f^Cv/S-l). 

2.^-2).''  4'    f[3V2-log(l+V2)]. 


380  ANSWERS 

5.    (a)  2irb(b  + - — cos-1- 

V        Va2-&2  a 


(6)  2 ^  +  -Jgg-  log  r«  +  vtf'-y-|. 

Va2  -  b2        La  -  Va2  -  62J 
«•  ¥«2-  7.    («)  tiWoH^;        9        ¥»*• 

8.   4*-d*  (/S)  7ra\/«2  +  &2.        10.    4  7r2aA:;  2w2a2k. 

11.    2  7ra2(3  sin  f i  —  3  ii  cos  <i  —  t\2  sin  £i). 

Pages  316-320.     Art.  155 


2. 

p  =  ea     . 

4.    p  =  —  a  cos  0  +  C. 

5. 

y*=  ax2  +  6. 

6.   pn  =  c  sin  n0.     Straight  line.     Cardioid. 

7. 

y  =  eax+c. 

8.     p: 

-h). 

_cea         9     ™1       10.    2  7ra2.        11.    a  log -2. 
3                                                    */i 

13. 

Va2  +  b2(t2 

14.  22.7  lb.         15.  0.9627  lb.        16.   4.4312. 

17. 

h  =  5.28  in. 

18. 

h  =  43.17  lb. 

Pages  323-324.     Art.  156 

3.    xy  =  ay2  +  by  —  \.       4.    y  =  kx  (log  x  -  1)  +  ax  +  b.       5.   T43  &«£. 

6,  y  =  0.0002  x3  +  0.0036z  +  1.12,  slope  =  0.0006  a2  +  0.0036  ;  a =20, 
y  =  2.792,  slope  =  0.2436. 

7.  1000  y=  -0.046  x3  +  0.75  x2  -  2  05  x  +  40;  1000?/=  ^  a8  -0.0575  x2 
-9.7x+117.91§. 

Page  328.     Art.  158 

1.  xy+C.  3.    Impossible.  5.    x3  +  y3  —  3  axy  +  C. 

2.  -  cos  a  cos  y  +  C.       4.    log-+C.  6.   tan-^+C. 

y  y 

7.  |  a3  +  x2?/  +  5  x  +  £  ?/3  -  \y2  +  <7. 

8.  £  x5  +  xy4  +  $  x3  -  xy2  +  |  y2  -  \  y*  +  2  ?/  +  (7. 

Page  330.     Art.  160 

1.  — =■  3.    J.  5.    |  a".  7-    «• 

2\/2  Z  3  2 

2.  4.  4.    1.  6.    6&3.  8.    J  log  2. 


ANSWERS  381 


Pages  332-334.     Art.  161 


6.    &.  c      Srfl; 


■Kd 


2 


2.  64.  4.    4  7T  -  8. 

3.  4£.  5.   1.  '8  9.    a2. 

/•y   /»a  /"a  /*7r  /*ir  /*o(l  — cos  0) 

10.    2l"f  pdpdtf;2j     I     _Vi    p.pfZ^p;2l     |  prfpdfl. 

JO   Ja(l-co8  0)  JO   Jcos      {}-*)  JnJa 

a  2 

1L   20.  12.   §JH— 2V3. 

o 

Pages  336-337.     Art.  162 

3  3  *  2  ^ 

Page  340.     Art.  163 

1.    62251b.     2.  1120.51b.     3.  9337.5  tons.     4.  66.41b.     5.    15.645  tons. 

Pages  343-345.     Art.  164 

i.  (t,M). 

2   (fcfA).  4.  fo,o,^-cV 

3.   x=8(*+a>g,y  =  g  =  0.  V  8; 

4(&  +  2a)'^ 


-     3|"(a  +  ft)4 a4"!  from  center  of  sphere. 

'   8|_(a  +  ft)3-a3J 
3r7(r  +  «)4-6^-|  from     herical 
8L7(r  +  a)3-6^J 


center. 


g    af  +  2flW  +  a^  above  tne  base. 
2(ab  +  a'b') 


o 
9.    from  the  vertex. 


4  a  sin  T 

~3 

2(r2  —  a'2^ 
10. from  center  of  circle. 

3  (V2  cos-1  --aVr2-  a2  ) 

»  (&  £)•  1S  (¥•  t)- 

10     /      2fl  2  6      \  «     /3a     36     3c\ 

12     [S(^Y)    '      3(7325)-  15'    U  '     8  '     8  j 


382  .  ANSWERS 

16  (*u     l6-^      16  c\ 
V15'    157r'    15  tt/ 

17  (<L*     8a     3«\.  -      -  =  ^  =  3r(«  +  ft)*-g*-i 
'    \  8  '     8  '     8  /'  8L(a  +  ^)3-a3J 

18.  x  =  y  =  z  =  —  - 

5 

19.  (0,  0,  —  j,  the  base  of  the  cone  being  in  xy-pl&ne.        20.    (0,  0,  \) 


21.    (a)  a  cos2  0  ;    (6)  a  (  1  H — ]  both  measured  from  the  vertex, 

\        1  +  cos2  d ) 

Pages  347-348.     Art.  165 

1.  «^(&2  +  c2).  6.    !!^(3«2  +  4^2). 

o  12 

2.  ^  ira^h.  ^     Tra&fr  /g  &2    i    #2) 

3.  7^(3a2  +  /l2)< 

12  8.    *£;    ^[(r  +  fc).-,*]. 

4     ttoH  15         15 

10     '  g  8  7IT"+5 

5.   \  Trabh(a2  +  tf) .  '    3(n  +  5)' 

10.  ia5.        11.  ^ira(2?*-r*).        12.  ^a36.        13.  \bh*.        14.  iira* 


INDEX 


(The  numbers  refer  to  pages) 


Absolute  value,  134. 
Absolutely     convergent, 

134. 
Acceleration,  77. 
Actual  velocity,  68. 
Approximate        integra- 
tion, 292. 
Arc,  length  of,  306. 
Area,  by  double  integra- 
tion, 330. 
derivative  of,  79. 
formula  for,  273,  280. 
in    polar    coordinates, 

291. 
in  rectangular  coordi- 
nates, 273. 
Asymptotes,  96. 
Average  curvature,  176. 

Bending,     direction     of, 
90. 

Cardioid ,  area  of,  292, 333. 
Catenary,  180,  316. 
length  of  arc,  309. 
volume  of  revolution, 
312. 
Catenoid,  313. 
Cauchy's  form  of  remain- 
der, 153. 
Center  of  curvature,  172. 
Center  of  gravity,  340. 
Change  of  variable,  124. 
Circle,    area    by    double 
integration,  331. 
of  curvature,  172. 
Cissoid,  180. 

area  of,  289. 
Component  velocity,  70. 
Concave,  89. 

toward  axis,  94. 
Conditionally       conver- 
gent, 134. 
Conditions    for  contact, 
168. 


Conjugate  point,  197. 
Conoid,  301. 
Constant,  15. 

factor,  25,  213. 

of  integration,  214. 
Contact,  167. 

of  odd  and  even  order, 
169. 
Continuity,  19,  109. 
Continuous  function,  19. 
Convergence,  132. 
Convex,  94. 

to  the  axis,  94. 
Critical  values,  55. 
Cubical  parabola,  281. 
Cusp,  194. 
Cycloid,  length  of,  311. 

surface  of  revolution, 
315. 

Decreasing  function,  51. 
Definite  integral,  270. 

geometric  meaning  of, 
273. 

multiple  integral,  329. 
Dependent  variable,  15. 
Derivative,  21,  22. 

of  arc,  79. 

of  area,  78. 

of  surface,  81. 

of  volume,  81. 
Determinate  value,  158. 
Development,  131,  160. 
Differentials,  74,  210. 

integration  of,  326. 

total,  112. 
Differentiation,  23. 

of   elementary    forms, 
44,  45. 
Direction  of    curvature, 

175. 
Discontinuous    function, 

20. 
Divergent  series,  132. 
Duhamel's  theorem,  348. 
383 


Ellipse,  area  of,  281. 

evolute  of,  190,  309. 

parametric  form,  290. 
Ellipsoid,  volume,  299. 
Envelope,  200. 
Epicycloid,  length  of,  311. 
Equiangular  spiral,  316. 
Evaluation,  160,  165. 
Evolute,  182. 

of  ellipse,  190,  183,  291. 

of  parabola,  183. 
Expansion  of  functions, 

131. 
Exterior  rectangles,  269. 

Family  of  curves,  200. 
Formula  for  integration 

by  parts,  216. 
Formulas  of  differentia- 
tion, 44,  45. 
of  integration,  211,  224. 
of  reduction,  229,  262. 
Function,  15. 

Helix,  318. 

Hyperbolic  branches,  95. 

spiral,  area  of,  292. 
Hypocycloid,  area  of,  290. 
length  of  arc  of,  308, 

311. 
volume    of    revolution 
of,  305. 

Implicit  function,  33. 
Impossibility    of    reduc- 
tion, 232. 
Increasing  function,  51. 
Increment,  21. 
Independent  variable,  15. 
Indeterminate  form,  157. 
Infinite,  20. 

Infinite  limits  of  integra- 
tion, 287. 
ordinates,  97. 
Infinitesimal,  350. 


384 


INDEX 


Integral,  209. 

definite,  270. 

double,  330. 

multiple,  329. 

of  sum,  212. 

triple,  328,  334. 
Integration,  209. 

by  inspection,  211. 

by  parts,  216. 

by  rationalization,  248. 

by  substitution,  219. 

formulas  of,  211,  224. 

of    rational   fractions, 
238. 

of    total    differential, 
326. 

successive,  321. 

summation,  268. 
Interior  rectangles,  269. 
Interval  of  convergence, 

138. 
Involute,  187. 

of  circle,  311. 

Lagrange's   form   of  re- 
mainder, 152. 
Lemniscate,      area      of, 

334. 
Length  of  arc,  306. 
of  e volute,  185. 
of  space  curve,  318. 
polar  coordinates,  309. 
rectangular        coordi- 
nates, 306. 
Limit,  15. 
change  of,   in  definite 
integral,  276. 
Limits,  infinite,  for  defi- 
nite integral,  287. 
Liquid  pressure,  338. 
Logarithm,       derivative 

of,  34. 
Logarithmic  curve,  280. 
spiral,  length  of  arc, 
310. 

Maclaurin's  series,  141, 

276. 
Maximum,  53. 
Mean  value  theorem,  74, 

275. 


Measure    of   curvature, 

177. 
Minimum,  53. 
Moment  of  inertia,  346. 
Multiple  points,  193. 

Natural  logarithms,  36. 
Normal,  85. 
Notation  for  rates,  72. 

Oblique  asymptotes,  99. 
Order  of  contact,  167. 

of  differentiation,  122. 

of  infinitesimal,  350. 
Osculating  circle,  172. 
Osgood,  132. 

Parabola,  108,  89,  171. 

semi-cubical,  308. 
Parabolic  branches,  95. 
Paraboloid,  314. 
Parallel  curves,  187. 
Parameter,  201. 
Partial  derivative,  110. 
Point  of  inflexion,  90. 
Polar  coordinates,  104. 

subnormal,  106. 

subtangent,  106. 
Problem   of    differential 
calculus,  21. 

of    integral    calculus, 
209. 

Radius     of     curvature, 
172. 

Rates,  68. 

Rational  fractions,  inte- 
gration of,  238. 

Rationalization,  248,  249. 

Rectangles,  exterior  and 
interior,  269. 

Reduction,  cases  of  im- 
possibility of,  236. 
formulas,  229,  262. 

Remainder,  150. 

Rolle's  theorem,  150. 

Simpson's  rule,  294. 
Singular  point,  191. 
Slope,  16. 
Solid  of  revolution,  81. 


Sphere,  volume'  by  triple 

integration,  336,. 
Spheroid,  oblate,  305, 314. 

prolate,  314. 
Spiral,    of    Archimedes, 
107. 

equiangular,   108s   316, 
310. 

hyperbolic,  292. 

logarithmic,  292. 
Standard  forms,  211, 224. 
Stationary  tangent,  90. 
Steps  in   differentiation, 

22. 
Stirling,  141. 
Subnormal,  86. 
Subtangent,  86. 
Summation,  268. 
Surface  of  revolution,  81. 

area  of,  312. 

Tacnode,  194. 
Tangent,  85. 
Tangent  plane,  119. 
Taylor,  141. 
Taylor's  series,  148. 
Tests    for    convergence, 

133. 
Total  curvature,  176. 

differential,  112. 
Tractrix,  315. 
length  of,  318. 
surface    of    revolution 

of,  318. 
volume    of   revolution 
of,  318. 
Transcendental        func- 
tions, 34. 
Trapezoidal  rule,  292. 
Trigonometric  functions, 
integration  of,  255. 

Variable,  15. 

Volume  of  solid*  of  revo- 
lution, 302. 

Volumes  by  triple  inte- 
gration, 334. 

Witch,  area  of,  286. 
volume    of    revolution 
of,  305. 


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